Suppose I have a Banach space $V$ and a set $A \subseteq V$ such that for all $\epsilon > 0$ there exists $v$ such that $A \subseteq \overline{B}(v, r + \epsilon)$. Does there exist $c$ such that $A \subseteq \overline{B}(c, r)$?

The answer is clearly yes for finite dimensional normed spaces: Define $T_\epsilon = \bigcap_{a \in A} \overline{B}(a, r + \epsilon)$. The $T_\epsilon$ form a chain of closed sets and for $\epsilon > 0$ are non-empty, so have the finite intersection property. Thus when $V$ is finite dimensional they have non-empty intersection, and any element of the intersection works as $c$.

For more general Banach spaces I feel like you should be able to choose a cauchy sequence $x_n$ such that $x_n \in T_{\epsilon_n}$ with $\epsilon_n \to 0$, but I can't seem to make it work.

Note that an arbitrary choice of $x_n \in T_{\epsilon_n}$ can't be guaranteed to be Cauchy: If $V$ is $l^\infty$ and $A = \{ x : x_0 = 0, ||x|| \leq 1 \}$ then diam$(T_\epsilon) \geq 2$ because you can choose $c_0$ arbitrarily in $[-1, 1]$

Note also that the assumption of $V$ a Banach space is essential: If $V$ is not Banach and $c$ is an element of the completion which is not in $V$ then $A = \overline{B}(c, 1) \cap V$ has no center.