Prove A is open if and only if w+A is open, A is closed if and only if w+A is closed.

The function $\phi(x) = x+w$ is a homeomorphism (continuous with a continuous inverse) hence it maps open sets to open sets and closed sets to closed sets (since closed sets are the complements of open sets).

Note that $w+A = \phi(A)$.

For part (a), it suffices to show that $B_r(w+a) = w + B_r(a)$. But this is easy since $$x\in B_r(w+a) \iff ||x-(w+a)||<r$$ $$\iff||(x-w)-a||<r$$ $$\iff x-w\in B_r(a)$$ $$\iff x\in w+B_r(a)$$

For part (b), just note that any set $S$ is closed iff $S^c$ is open. So, $$A\text{ closed } \iff A^c \text{ open }$$ $$\iff w+A^c \text{ open }$$ $$\iff (w+A)^c \text{ open }$$ $$\iff w+A\text{ closed } $$