Baby Rudin's proof that non-empty Perfect subsets of $\mathbb{R}^k$ are uncountable

Question

I'm translating from the french version but I saw a copy of the proof from the english edition and the french translation seems identical. Anyhow I think I have the terminology right.

Suppose $\emptyset \ne P \subset \mathbb R^k$ is a perfect set and assume it is countable (it being infinite is easy to prove). Let the elements of $P$ be numbered $x_1,x_2,\dots$

We'll construct open balls $B_n$ of center $x_n$ in the following way. (Notice here that $B_n$ is already stated to have center $x_n$)

Let $B_1$ be any open ball, in particular since $x_1 \in B_1,\ \overline{B_1} \cap P \ne \emptyset$. Moreover $x_1$ is an accumulation point of $P$ ($P$ is perfect after all) and so $B_1$ must contain an infinite number of points of $P$.

Now suppose $B_n$ is well defined such that $B_n \cap P$ is non-empty, since every point of $P$ is an accumulation point, there is an open ball $B_{n+1}$ such that $(i)\ \overline{B_{n+1}} \subset B_n,\ (2)\ x_n \notin \overline{B_{n+1}},\ (iii)\ \overline{B_{n+1}} \cap P$ is not empty.

Now this is where this whole proof grinds me to a halt, $B_{n+1}$ was already stated to have center $x_{n+1}$, but it is not clear at all that $x_{n+1}$ was in $B_n$ in the first place, it could have been anywhere for that matter. I read a similar proof that was picking the center of $B_{n+1}$ after $B_n$ is defined which makes more sense since $x_n$ is an accumulation point of $P$. However if you do that, there's no way to guarantee you've picked every point of $P$ making the next step of the argument incorrect:

Let $K_n = \overline{B_n} \cap P$, since $P$ is perfect, it is closed, so is $\overline{B_n}$, moreover it is bounded and so $K_n$ is compact. Therefore $\bigcap_{1\leq j \leq n} K_j$ is a non-empty compact subset of a metric space for all $n$. However, for all $n$, $x_n \notin \bigcap_{1 \leq j \leq n+1} K_j$, hence for all $n$, $x_n \notin \bigcap_{j \in \mathbb N} K_j$. Therefore $\bigcap_{j \in \mathbb N} K_j \subset P$ contains none of the points of $P$. This is impossible since we've shown before that such a set must be non-empty.

So in my view, either the set $x_1,x_2,\dots$ is predefined and it is not possible to guaranted $(i)\ \overline{B_{n+1}} \subset B_n$, or the last part of the argument is incorrect and it can't be guaranteed to have excluded every point of $P$.

I'm not familiar with the Axiom of Choice/Dependent Choice, I've been acquainted with Zorn's Lemma.

Answer 1

I believe this answers my question:

Need help in Understanding the Proof of Theorem 2.43 on Perfect Sets in Baby Rudin.

But don't know how to resolve my question. In any event, this seems to show the presentation in Rudin quite shoddy.

edit: The reason why the above link answers my question is that if say $x_2 \notin B_1$, then we can pick any other point $x' \in B_1\backslash\{x_1\}$ with $B_2 = B(x',||x'-x_1||)$ and now $x_2 \notin B_2 \subset B_1$.

This gives us the expected property that no points of $P$ is in $\bigcap_{n \geq 1} K_n$.

2nd edit: The reason the french version is misleading is that it states the $x_n$ must be the center of $B_n$, what must actually be understood is that the denumerable sequence of points $x_1,x_2,\dots$ must indeed have been established and as long as we make a choice of open ball $B_{n+1} \subset B_n$ we can guarantee that $x_n$ isn't in it regardless of its center thereby guaranteeing $P$ is disjoint from $\bigcap_{n \geq 1} K_n \subset P$ leading to the contradiction.