Show that there exists a point $a \in A$. Prove or disprove.
Suppose $A,B$ are subsets of $\mathbb{C}$ with $A$ compact. Show that there exists a point $a \in A$ such that: for all $x \in A$ and $y \in B$, there exists $b \in B$ such that $|a - b| \le |x - y|$.
Any help to point me in the right direction would be highly appreciated.
 A: *

*Let's introduce function $\rho_B(x) = \inf_{b\in B}|x-b|$. Since $|x-b|\ge 0$, $\rho_B(x)\ge0$ too. One can show that this function is continuous.

*On a compact set $A$, continuous function $\rho_B(x)$ reaches its minimum. So there exists such point $a\in A$, that
$$ \rho_B(a)=\min_{x\in A} \rho_B(x)$$
Here goes the tricky part. It can be the case, that there are number of possibilities to choose $a$. For some choices the infimum of $\rho_B$ will be reachable (some point $b_a\in B$), and for some not. We always choose $a$ with reachable infimum if such $a$ exists. 
On the picture below, both $\rho_B(a_1)=\rho_B(a_2)=\min_{x\in A}\rho_B(x)$. However, for $a_1$ the infimum is unreachable, while for $a_2$ it is otherwise. So we will choose $a_2$ in this case.



*If we have reachable infimum $b_a$, then we will always choose it as $b$. For every $x\in A$ and $y\in B$:
$$
|a-b| \le |x-y|
$$
from definitions of $\inf$ and $\min$.

*If there is no $a$ with reachable infimum, we can take a sequence $b_n$, so
$|a-b_n|\to\rho_B(a)$. For every $x$ and $y$, $|x-y| > \rho_B(a)$ (if it's $=$, then $x$ has a reachable infimum and we should have chosen $x$ as a). So there exists such $b_n$, that
$$
|a-b_n|-\rho_B(a) < |x-y| - \rho_B(a), \qquad |a-b_n| < |x-y|
$$
P.S. It's an outline of a proof. You may need to prove all the steps properly. Please ask if something is not clear.
