For the first one, I think this might work:
Suppose that there exist two distinct balls with such property, lets say, $B(x,r)$ and $B(y,r)$. Take the balls $B(x,\delta)$ and $B(y,\delta)$ with $\delta$ small such that $B(x,\delta)\cap B(y,\delta)\neq\emptyset$
Now you vary $\delta$ to the first value (lets call it $s$) where $B(x,\delta)$ and $B(y,\delta)$ intersects on only one point (this is possible $H$ is a Hilbert space and hence uniformly convex). Let this point be $z$.
Can you see now that it's possible to find $r' such that the ball $B(z,r)$ cover $A$?
For the second one:
If $A$ is convex and closed the problem $\min\{\|u-v\|,\ v\in A\}$
is solvable for every $u\in\mathbb{R}^{n}$ and have just one solution that we gonna call $P_{A}(u)$ (projection of $u$ in $A$) and $P_{A}(u)\in A$.
Suppose that the circumcenter $u$ of $A$ is not in $A$. So the number $t=\|P_{A}(u)-u\|>0$ is well defined. This means that the ball $B(u,t)$ just intersects $A$ in $P_{A}(u)$.
On the other hand the ball $B(P_{A}(u),t)$ contains "much more" (try to give a precise meaning for this) elements of $A$, hence, you can find $r' such that the ball $B(P_{A}(u),r')$ cover $A$.