$\newcommand{\d}[1]{\mathtt{\large{\underline{#1}}}}$I will cover, for the sake of others who may not have understood what you could from the Professor's e-mail, the entire trail and logic of reasoning of the Professor. Therefore, while OP will know some of the material here, it is important everybody can read through the entire mail.
The setting
The basic idea is that of serving agents. You have agents, and they all require a certain facility. For example , people live in a certain locality and you want to place a grocery store to satisfy their food requirements. The way you place the store would be based on how far you are from each of them. Each person will associate a cost , which is the amount of effort they must expend to reach the store, and you want to satisfy everybody to the best of your ability. This is modeled in the following way mathematically.
We have a subset $S \subset [0,1]$. There is meant to be an agent or customer at $S$ i.e. a person who needs to be served. There is a single facility or grocery store to satisfy everybody, located at a $y \in [0,1]$. Then we talk about the cost on the entire locality of having the store at this place.
The notion used here is the notion of the $l_1$ distance. So for each point $x \in S$, the cost of having the facility at $y$ is $|y-x|$, which you can think of as the distance to travel from $y$ to $x$. Then we sum over $x \in S$ to get the function $f(y) = \sum_{x \in S} |y-x|$.
$f$ is then the cost to the locality of having the facility at $y$. The reason why $f$ can be interpreted as such , I have already expounded on.
Naturally, the question to be asked is : does $\min_{y \in [0,1]} f(y)$ exist? Is it unique? Can we find it(or one of the minima) in a certain way (maybe using an algorithm)? Of course you wish to answer these questions so you can place the locality in a place obtained by using such an algorithm.
The questions addressed of $\color{blue}{\text{existence}}$, $\color{red}{\mathrm{uniqueness}}$ and $\color{green}{\mathrm{construction}}$ are all part of the general notion of well-formulation or well-posedness of a problem in the field of optimal control theory. Indeed , the specific question addressed by the Professor is a very minimal example in the word of optimal control.
On the well-definition of $f$
First of all, $f$ is defined as the sum over all $x\in S$ of non-negative quantities. It can be noticed that if any reasonable notion of summation allowed $S$ to be uncountable , then $f$ is obsolete, in the sense that it is either zero or infinite everywhere. Thus, such a situation is not interesting. You can see here for the details on that.
We therefore restrict ourselves to $S$ being either finite or countable. However, it seems that all assertions have been made for the countable case. With this, let us prove the first assertion generalized :
If $S$ is countable and $y$ is not a limit point of $S$ then $f(y) = \infty$. On the other hand, if $S$ is finite then $f(y)< \infty$ for all $y \in [0,1]$.
Proof : The easy case first. If $|S| = m< \infty$ then $|x-y| \leq 1$ for all $x \in S, y \in [0,1]$ so $f(y) \leq m < \infty$ for any $y \in [0,1]$.
The harder case now : if $|S| = \infty$ and $y$ is not a limit point of $S$, then in particular there exists an $\epsilon>0$ and points such that $|x - y| > \epsilon$ for all $x \in S$ (using the non-finiteness of $S$ here). But then $f(y)\geq \sum_{i=1}^n |x_i - y| \geq n \epsilon$ (ordering the points in $S$ like $x_1,x_2,...$ arbitrarily) for all $n$ so $f(y) = \infty$.
This means that $f$ doesn't make sense whenever $y$ is not a limit point of $S$. But what if $S$ had no limit points? The answer to that question is addressed by the following result :
Every infinite subset of a compact subset of $\mathbb R$ has a limit point.
This is a consequence of the Bolzano-Weierstrass theorem. Thus, we can rest assured that $S$ will certainly have limit points, and $f$ can possibly be finite at one of these limit points, so we at least have a problem to work with. We may then seek to look at the behaviour of these points.
Looking at the limit points
The problem with more than one limit point, is that if points are close to one limit point they are not close to the other, so that makes $f$ infinite at both these limit points. However, even if $y$ is a limit point of $S$, the rate at which points in $S$ approach $y$ may not be good enough for summation. It is illustrative to explain this with examples.
Let $S = \{1,\frac 12, \frac 13 , ...\}$ (not including $0$). Then $0$ is the only limit point of $S$ (easy to see, it is a monotonically decreasing sequence and then one can use the Archimedean axiom), so we can look at $f(0) = \sum_{n \in \mathbb N} \frac 1n$. This is the famous harmonic series, which is known to be infinite. Thus $f(0) = \infty$. In particular, for $S = \{1,\frac 12,...\}$, the problem of finding a facility is obsolete!
We may consider a slight modification : let $S = \{1 , \frac 14 , \frac 19 ,...\}$ denote the squares of the reciprocals of the natural numbers. Once again, $0$ is the only limit point. However, note that $f(0) = \frac 1{n^2} < \infty$ (a well-known result). Thus, the problem is well-defined, and furthermore we have that $0$ must be the unique minimizer, since $f$ is infinite at all other points.
For some situation involving multiple limit points, let $$S = \left\{\frac 14 , \frac 19, \frac 1{16},...\right\} \cup \left\{1 - \frac 1{8} , 1 - \frac 1{27} , 1 - \frac 1{64} ,...\right\}$$, so it's the squares of the reciprocals except $1$, along with the cubes of the natural numbers except $1$, subtracted from $1$. The deliberate construction of the sequence tells us that $0,1$ are the only limit points. However, one can see that $f(1) = f(0)=\infty$, because although the first part of the definition of $S$ converges to $0$, it goes far away from $1$, so $f(1) = \infty$. Similar the second part converges to $1$ so goes far away from $0$ so $f(0) = \infty$. So the problem becomes obsolete.
We can now prove using the examples, the following result for $S$ countable:
For $S$ countable, there is at most point $y \in [0,1]$ such that $f(y) < \infty$. That point, if it exists, is the unique limit point of $S$.
Proof : Suppose that $f(y_1)<\infty$ and $f(y_2) < \infty$ for $y_1 \neq y_2$. Then they are limit points from the explanation in the previous section, so there exist subsequences $x_1,x_2,...$ and $z_1,z_2,...$ of $S$ such that $|x_i - y_1|< \frac{|y_1 - y_2|}{2}$ and $|z_i - y_2| < \frac{|y_1 - y_2|}{2}$ for all $i$. By the triangle inequality, $|x_i - y_2| > \frac{|y_1 - y_2|}{2}$ and $|z_i - y_1|> \frac {|y_1-y_2|}2$ for all $i$. It follows that $f(y_1) = f(y_2) = \infty$.
Which means that $f(y)<\infty$ at possibly only one point, which must a limit point as well. There cannot be any other limit point by the previous argument so $y$ is the unique limit point of $S$.
Now, given $S$ we can assume that $y$ is the unique limit point of $S$ and talk about $f(y) = \sum_{x \in S} |x-y|$. However, as I illustrated in the first two examples above, the finiteness of this sum depends upon how fast $|x-y|$ approaches $0$ as $x$ approaches $y$.
The theory of convergence of series will dictate how fast it needs to be. Tests like the ratio test, comparison test etc. are the basic tools in this direction.
The notions of rate and order are important, because these dictate the long-term behaviour of how $|x-y|$ looks like for $x$ close to $y$, and thus determine the summability of $f(y)$. For example, if $S = \{\frac 2n\}$ then the rate of convergence to the limit point $0$ is $O(\frac 1n)$ which is too slow. However, take $S = \{\frac 1{2n^3} + \frac 1{2n^2}\}$ then the convergence to $0$ is $O(\frac 1{n^2})$ which is fast enough for convergence.
The example proposed by the OP is not a counterexample, but rather an example! Indeed, the shrinking of the harmonic series is too slow to allow convergence to happen, as Professor's point number $2$ asserts.
Finally, to address $S$ finite :
If $S$ is finite then $f(y)$ is a continuous function of $y$.
Proof : $f(y) = \sum_{x \in S} |x-y|$ is now a finite sum of continuous functions, hence continuous.
Thus $f$ attains a minimum on $[0,1]$, so the problem is well-defined! You can try to see when the minimum is unique (for example, $S$ consisting of one point?) by experimentation : it is not unique all the time!
On the construction part : if $S$ is infinite, then the question is of the location of the limit point of $S$, if it is unique, and if the rate of convergence is fast enough, which are all found analytically from a symbolic expression of $S$ (I actually don't know how $S$ is specified in finite terms if it is an infinite set , so I won't speak of the algorithmics). In the case of $S$ finite, the additional factor helping us is that $f$ is convex, whence we can apply a gradient descent or so to get a minimum.
We see that the problem is a bit obsolete, in the case of $S$ countable (for $S$ finite, it is a tractable problem, even if the solution is not unique). We allow more flexibility for $S$ countable, by changing the $f$.
A changed $f$ and the third point
We change our setup now. We have $m$ competing grocery stores , or facilities, and each person prefers the grocery store closest to him or her. That is , for a facility located at $y$, we look at all the $x$ such that $x$ is closer to $y$ than to any other facility, call that subset $S'$, and then $f(y) = \sum_{x \in S'} |x-y|$. So the cost of a facility doesn't take everyone into account now, but because of competition only those customers which are likely to be closest to it.
To be rigorous : Let $\{y_1,...,y_m\}$ be the locations of the facilities. For $y \in \{y_1,...,y_m\}$ let $S'_y = \{x \in [0,1] : |x-y| \leq |x-y_i| \text{ for all} i = 1,2,...,m\}$, and then define :
$$
f(y) = \sum_{x \in S'_y} |x-y|
$$
With this in mind, let us look at example $3$ that we wrote earlier. There, the cost was infinite.Now, with this new metric , this is not the case.
Indeed, $1$ and $0$ are the two limit points, but the sums for both are finite, because we aren't counting the points closer to $0$ in the sum for $1$, and vice-versa. In that case, we have under the new $f$ :
$$
f(0) = \frac 14 + \frac 19 + \frac 1{16} + ... , f(1) = \frac 1{8} + \frac 1{27} + ...
$$
and $f(1)<f(0)<\infty$ (To see $f(1)<f(0)$ we can use a pointwise minimization that $\frac 18<\frac 14 , \frac 1{27}<\frac 19$ etc. and sum up). So $1$ is a better facility than $0$ under the cost structure now imposed!
We can instantly see how the change of $f$ has improved the quality of the problem we seek to solve. This is innovative mathematics in a nutshell : changing the parameters of the problem to produce and give meaning to new unexplored situations.
With this cost structure, we can talk about the total cost $F(y_1,...,y_m) = \sum_{i=1}^m \sum_{x \in S'_{y_m}} |x-y_m|$ where $F : [0,1]^m \to [0,\infty]$. So you place each of the facilities, and total up the costs for each agent. You want to minimize $F$ , naturally, so you can place your facilities there. The question is : when is $F$ finite?
The answer to this question is given wrongly by the Professor. The correct statement is this:
Let $F$ be given as previously. Then $F$ is finite if and only if for each $i$ we have either $S'_{y_i}$ is finite or $y_i$ is a limit point of $S'_{y_i}$ (and hence of $S$) and the decay $|x-y_i|$ is fast enough.
Proof : Let any $y_i$ not be a limit point of $S$. It is then not a limit point of $S'_{y_i}$ either. Suppose that $S'_{y_i}$ is finite, then the resulting sum $\sum_{x \in S'_{y_i}} |x-y_i|$ is finite even if $y_i$ is not a limit point of $S$!
However, if $S'_{y_i}$ is infinite, then we can use the $\epsilon$ method we used earlier to get a contradiction. Thus $y_i$ must be a limit point of $S'_{y_i}$.
Now the statement is near obvious, a finite sum of quantities is finite if and only if each one is finite, so $F$ is finite if and only if each summand is finite.
Let me give an example to clarify this. Let $S = \{\frac 14, \frac 19,\frac 16,...\} \cup \{0.85,0.95,0.99\}$. Let's say we want two facilities. I can place one facility at $0$ to accommodate for the first sequence. To acommodate the finitely many leftover points, I can put the second facility anywhere close to them to retain finiteness of $S'$ of this point! For example, $F(0,0.75) < \infty$ although $0.75$ is not a limit point of $S$. In fact, $F(0,a)<\infty$ for any $a$, since any other $a$ is not a limit point so the set $S'$ for it will be finite and hence sumability will follow.
For finite $S$, things are clear anyway. Construction and uniqueness arguments are skipped, these are similar to the previous situation.
Conclusion
In conclusion, we have seen two problems in optimal control and studied their well-definition and the existence of solutions to such problems. This serves as a primer for subsequent theory in optimal control, which includes :
Convex analysis and optimization (which is interesting because the individual cost functions were convex in the first part).
Using theorems to assert existence (we had existence here from the theorem that a continuous function on a closed and bounded interval attains its minimum : generalizations exist, but the more common tool seen is the Banach fixed point theorem and cousins)
Innovating our notion of distances to allow for flourishing of existing theory (the second problem was much richer than the first, and generalized the first problem in fact. We can also attempt to perform an $l_2$ optimization by taking $(\sum_{x \in S} (y-x)^2)^{\frac 12}$ in place of the $l_1$ sum, which leads to very interesting mathematics because there's an explicit formula for a minimizer in this case).
