= Solution
Let the radius be $r>0$. One inclusion follows from (a). For the reverse, if $\|x-c\|\le r$, set $x_n=c+(1-1/n)(x-c)$ for $n\ge2$. Then $\|x_n-c\|<r$ and $x_n\to x$, so every closed-ball point is in the <closure> of the <open ball>. Hence
$$
\boxed{\operatorname{cl}B(c,r)=\overline B(c,r)\quad(r>0).}
$$
The 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.
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