TITLE: Covering a set with $N$ balls of common minimal(!) radius (Existence of such a covering)
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: I don't feel too bad posting a "proof by references" since I worked on this problem for quite a while but failed, before realizing the $N=1$ case seems pretty google-able. In any event, it seems Konyagin proved that any non-reflexive Banach space admits an equivalent norm and some 3 points so that there is no smallest ball with respect to the equivalent norm containing those 3 points. A good buzzword is "Chebyshev center". Below is a link to a strengthening of Konyagin's result (and a mention of Konyagin's result). I couldn't find a free version of Konyagin's paper online.
https://dml.cz/bitstream/handle/10338.dmlcz/119230/CommentatMathUnivCarolRetro_42-2001-1_11.pdf