Theorem 19.7.4 (Blaschke selection theorem)BA1F1C1CF20F

Proof. Fix $R>0$ and let $\mathcal K(R)$ be the nonempty compact convex subsets of the closed ball $\overline B(0,R)$. For $K\in\mathcal K(R)$ its support function $h_K$ on the unit sphere is bounded by $R$ in absolute value and is $R$-Lipschitz. A diagonal selection on a countable dense subset of the sphere, followed by the common Lipschitz bound, gives a uniformly convergent subsequence $h_{K_j}\to h$ from every sequence.

Let $K=\{x:\langle u,x\rangle\leq h(u)\text{ for all unit }u\}$. This is closed, convex and contained in $\overline B(0,R)$. It is nonempty: select $x_j\in K_j$ and pass to a convergent subsequence; the limit satisfies all the inequalities. For every fixed $u$, select $x_j\in K_j$ maximizing $\langle u,x\rangle$ and pass to a subsequence. Its limit lies in $K$ and attains $h(u)$, so $h=h_K$.

For compact convex sets, $d_{\mathrm{Haus}}(K,L)=\sup_{|u|=1}|h_K(u)-h_L(u)|$. One inequality follows by approximating each point of either body within its Hausdorff distance. For the other, an upper bound $\delta$ on the support-function difference gives $K\subset L+\delta\overline B$ and the reverse inclusion by separation of a point from a closed convex set. Hence $K_j\to K$ in Hausdorff distance. We have proved compactness of $\mathcal K(R)$. Every sufficiently small closed Hausdorff ball about a given body lies in some $\mathcal K(R)$ and is closed there, so is compact. In dimension zero the space has just one member.