Compact convex sets as intersection of balls 
Prove that any compact convex set $K\subset \mathbb{R}^d$ is the intersection
of some balls.

Hello, I saw this problem during my Geometry class, but I don't know how to prove it. Can someone help me?

Let $K$ be a compact convex set in $\mathbb{R}^d$. For each $x \notin K$, let $R_x = \inf\{r: K \subset \overline{ B(x, r) }\}$. This is well-defined because $K$ is compact and hence bounded. Clearly $$K \subset \bigcap_{x \notin K} \bigcap_{r > R_x} \overline{B(x, r)} = S.$$ For $x \notin K$ let $w_x$ be the unique point in $K$ such that $|x-w_x|$ is minimum; the existence and uniqueness of $w_x$ follows from the fact that $K$ is closed and convex and $\mathbb{R}^d$ is a Hilbert space (see for example this handout).
For the reverse inclusion: let $x \notin K$ and consider the points $x_t = x + t(w_x - x)$, i.e., the points on the ray in the direction of $x$ to $w_x$. The hyperplane perpendicular to this ray and passing through $x$ divides $\mathbb{R}^d$ into $2$ halves, and $K$ is strictly contained in the interior of the half-space on the same side as the ray. For $t$ large enough, the closed ball $\overline{B(x_t, |x_t - x|)}$ must contain $K$ in its interior (since in the limit $t \to \infty$ this ball will contain the entire half-space). Moving this ball by a small amount along the ray will give a ball that does not contain $x$ and still contains $K$. This ball is in the intersection above, so $x \notin S$.
