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 $\lVert c_n\rVert=1\not\to0$ and $c\not\in W$, contradicting the assumption.
QED