 Solution

Let the radius be $r>0$. One inclusion follows from (a). For the reverse, if $\Vert x-c\Vert \le r$, set $x_n=c+(1-1/n)(x-c)$ for $n\ge 2$. Then $\Vert x_n-c\Vert <r$ and $x_n\to x$, so every closed-ball point is in the closure of the open ball. HenceThe positive-radius convention matters: if radius zero is allowed, the open ball is empty whereas the closed ball is $\{c\}$, so the asserted equality fails at that degenerate radius.