Let $X$ be a Banach space, and denote by $B_r (x)$ the closed ball of radius $r > 0$ around $x \in X$. Furthermore, let $A \subset X$ be compact and $N \in \Bbb{N}$. I am interested in "optimally" covering $A$ by $N$ balls, i.e,. with as small radius as possible. More precisely, define $$ r_0 := \inf \Big\{ r > 0 \colon \exists \, x_1,\dots,x_N \in X \text{ such that } A \subset \bigcup_{i=1}^N B_r (x_i) \Big\}, $$ and assume that $r_0 > 0$.
I would like to know whether there necessarily exist $x_1,\dots,x_N \in X$ such that $A \subset \bigcup_{i=1}^N B_{r_0} (x_i)$. In other words, I would like to know if the infimum above is actually a minimum.
Note I am taking the $x_i$ from the "surrounding" space $X$, not from the compact set $A$.
I can prove the claim in case that $X$ is reflexive (even only assuming that $A$ is bounded), but I am not sure whether it is true for more general Banach spaces. I will give my proof for the reflexive case below, in case one can either generalize it, or use it to get an idea for a counterexample.
Proof for the reflexive case: Choose a sequence $r_n \to r_0$ such that for each $n$ there are $x_1^n,\dots,x_N^n \subset X$ satisfying $A \subset \bigcup_{i=1}^N B_{r_n}(x_i^n)$. If $B_{r_n} (x_i^n) \cap A = \emptyset$ for some $i,n$, replace $x_i^n$ by zero. Note that this retains the property $A \subset \bigcup_{i=1}^N B_{r_n}(x_i^n)$.
Since $A$ and the sequence $(r_n)_{n}$ are bounded, there is $R > 0$ such that $\| x \| \leq R$ and $r_n \leq R$ for all $x \in A$ and $n \in \Bbb{N}$: There are now two cases for each $i,n$: 1) There is some $x \in A \cap B_{r_n} (x_i^n)$, and hence $\| x_i^n \| \leq \| x_i^n - x \| + \| x \| \leq r_n + R \leq 2R$. 2) There is no $x \in A \cap B_{r_n} (x_i^n)$, and hence $x_i^n$, whence $\| x_i^n \| \leq 2R$.
Therefore, each of the sequences $(x_i^n)_{n \in \Bbb{N}} \subset X$ is bounded. Since $X$ is assumed to be reflexive, we can choose a common subsequence (which I will ignore in the notation below) such that $x_i^n \to x_i$ weakly for all $i = 1,\dots,N$.
Now, let $x \in A$ be arbitrary. For each $n \in \Bbb{N}$, there is $i_n \in \{1,\dots,N\}$ satisfying $\| x - x_i^n \| \leq r_n$. Next, there is $\ell \in \{1,\dots,N\}$ such that $i_n = \ell$ for infinitely many $n \in \Bbb{N}$, say for $n = n_m$ with $n_m \to \infty$. Since $x - x_i^{n_m} \to x - x_i$ weakly and since the norm is lower semicontinuous with respect to weak convergence, we see that $\| x - x_\ell \| \leq \liminf_{m \to \infty} \| x - x_i^{n_m} \| \leq \liminf_{m \to \infty} r_{n_m} = r_0$. Since this holds for any $x \in A$, we get $A \subset \bigcup_{\ell=1}^N B_{r_0} (x_\ell)$, as desired.
