In this paper we find a proof that for every continuous function $f:S^n\to\mathbb{R}$ there exist $n+1$ pairwise orthogonal points $x_0$ through $x_n$ such that $f(x_0)=f(x_1)=\cdots=f(x_n)$. They state that this proves the following: any convex body $B$ has a circumscribing hypercube, that is, a hypercube enclosing the convex body such that each face intersects the convex body, and give a reference to the proof of this by Kakutani. I presume that goes like this. We may define the continuous $f:S^n\to\mathbb{R}$ by $f(x)$ is maximum distance between the projections of two points in $B$ onto the span of $x$. Then it can be seen that $B$ is circumscribed by the figure enclosed by the planes given by these projections at the perpendicular $x$ with the same value. At these $x$, $f(x)\neq0$ as otherwise $B$ can contain at most one point, so the figure is a non-degenerate hypercube. Okay, so why do we we need the theorem to be about convex bodies? Doesn't everything I just said apply to any compact set with more than one point? I thought maybe $B$ and the hypercube must be tangent at a point on each face not contained in the edges of the hypercube, but this is necessary for a convex body given by two identical very pointy cones joined at their bases which any circumscribing hypercube touches only at two opposite vertices.
EDIT:
I swear to god I’m right on this. Here’s more than a sketch of a proof. Let $f:S^n\to\mathbb{R}$ be given by $f(x)=\max_{b,c\in B}\|\textrm{proj}_xb-\textrm{proj}_xc\|$. $f$ is well defined since unbounded projections or an unrealised supremum would contradict compact non-emptiness. $f$ is continuous. Since $\textrm{proj}_xb$ is jointly continuous in its arguments and $S^n\times B$ is compact, for all $x$, for all $\varepsilon>0$, there exists a neighbourhood $U$ of $x$ such that for all $u\in U$, and $b,c\in B$ the absolute difference between $\|\textrm{proj}_xb-\textrm{proj}_xc\|$ and $\|\textrm{proj}_ub-\textrm{proj}_uc\|$ is less than $\varepsilon$. Thus the maximum over $b$ and $c$ is within $\varepsilon$ of its value at $x$. By the theorem, there exist $x_0$ through $x_n$ such that $f(x_0)=f(x_1)=\cdots=f(x_n)$. Let $b_i$ and $c_i$ in $B$ achieve $f(x_i)$ at $x_i$. Then taking the hyperplanes $\{v\ |\ \textrm{proj}_{x_i}v=\textrm{proj}_{x_i}b_i\}$ and $\{v\ |\ \textrm{proj}_{x_i}v=\textrm{proj}_{x_i}c_i\}$, each pair has the same distance $f(x_i)$ and is perpendicular to to the other pairs of hyperplanes. Obviously $B$ intersects every pair and $B$ must be enclosed by every pair since otherwise we could take a point with projection on one of the hyperplanes and another with projection beyond the other to get a greater distance. If $f(x_i)=0$ then clearly $B$ only contains one point, a contradiction. So our hyperplanes define a hypercube.