I believe it's not true in general, but my counterexample is a little involved.  If you've the patience then follow me through...

Let $V=\bigoplus_{n=1}^\infty \ell^n_2$ where $\ell^p_2$ is $\mathbb{R}^2$ with
norm $\lVert\cdot\rVert_p$ (Note: $n$ is taking the role of $p$).  For $i\geq1$
and $j\in\{0,1\}$ we have $e_{i,j}$, the $j^{th}$ standard basis vector of
$\ell^i_2$ in $V$.

Give $V$ the norm $\lVert v\rVert=\sup_n\lVert v_n\rVert_n$.

Let $W=\{v\in V:\lVert v_n\rVert_n\to 0\}$.  I assert that $W$ is a Banach space.
Certainly every $e_{i,j}\in W$.

Let $A=\{e_{k,0}+e_{k,1}, e_{k,0}-e_{k,1}:k\geq 1\}$.

Fact: $r(A)\leq1$

Proof:

Let $c_N=\sum_{i=1}^n e_{i,0}$.  We wish to compute the distance of each point
of $A$ from $c_N$.

For $k\leq N$ we have
$\lVert c_N-e_{k,0}-e_{k,1}\rVert$
$=\lVert\sum_{i=1\ (i\not=k)}^Ne_{i,0}-e_{k,1}\rVert$
$=\sup\{\lVert e_{i,0}\rVert_i:i\leq N,i\not=k\}\cup\{\lVert-e_{k,1}\rVert_k\}$
$=1$
and similarly for $\lVert c_N-e_{k,0}+e_{k,1}\rVert$.

For $k>N$ we have
$\lVert c_N-e_{k,0}-e_{k,1}\rVert$
$=\max(\lVert c_N\rVert,\lVert e_{k,0}+e_{k,1}\rVert_k)$
$= \max(1,(1+1)^\frac{1}{k})$
$= 2^\frac{1}{k}$
$\leq 2^\frac{1}{N}$
and similarly for $\lVert c_N-e_{k,0}+e_{k,1}\rVert$.

Thus $A\subseteq \overline{B}(c_N,2^\frac{1}{N})$ and so $r(A)\leq2^\frac{1}{N}$.  Letting
$N\to\infty$ we have $r(A)\leq 1$.

QED

Fact: $A$ is not contained in a ball of radius $1$.

Proof:

Suppose $A\subseteq \overline{B}(c,1)$.  Then in particular for every $n$ we have
$\lVert c-e_{n,0}-e_{n,1}\rVert\leq 1$ and thus $\lVert c_n-e_{n,0}-e_{n,1}\rVert_n\leq 1$.  Similarly $\lVert c_n-e_{n,0}+e_{n,1}\rVert_n\leq 1$.

Simple consideration of $\ell^n_2$ shows that this implies $c_n=e_{n,0}$.
Thus $c_n\not\to0$ and $c\not\in W$, contradicting the assumption.

QED