 Solution

Take radius $r>0$. Part (a) gives one containment. For a point $x$ with $\Vert x-a\Vert \le r$, define $x_t=a+t(x-a)$ for $0\le t<1$. Then $\Vert x_t-a\Vert =t\Vert x-a\Vert <r$, while $x_t\to x$ as $t\uparrow 1$. Hence every point of the closed ball belongs to the closure of the open ball, proving equality. Positive radius is the standard open-ball convention; a zero-radius open ball would be empty.