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[This is an attempt to explain the details in Anton Petrunin's answer to this question, since the comments suggest that a number of people have found it hard to understand as it was written.]

[This is an attempt to explain the details in Anton Petrunin's answer to this question, since the comments suggest that a number of people have found it hard to understand as it was written.]

Let $C$ be the surface of the unit cube in $\mathbb{R}^d$ (for $d\geq 2$); when considered as a metric space, it is endowed with the distance-on-the-surface: so the point is to show that the distance between two antipodal points is $2$ (clearly it is at most $2$). Let $S$ be the ($(d-1)$-dimensional) sphere of radius $2/\pi$, also endowed with the intrinsic distance: certainly the distance between two antipodal points is $2$. So we are done if we can find a $1$-Lipschitz (="short", =which doesn't increase distances) map $h\colon C\to S$ mapping antipodal points to antipodal points. Note that this map does not have to be injective.

Let $C$ be the surface of the unit cube in $\mathbb{R}^d$ (for $d\geq 2$); when considered as a metric space, it is endowed with the distance-on-the-surface: so the goal is to show that the smallest possible distance between two antipodal points is $2$ (clearly it is at most $2$). Let $S$ be the ($(d-1)$-dimensional) sphere of radius $2/\pi$, also endowed with the intrinsic distance: certainly the distance between two antipodal points is $2$. So we are done if we can find a $1$-Lipschitz (="short", =which doesn't increase distances) map $h\colon C\to S$ mapping antipodal points to antipodal points. Note that this map does not have to be injective.

This map still needs to be modified a little: let $h\colon Q\to K$ be obtained by composing the nearest point projection $Q\to P$ (which makes sense since $P$ is convex compact) with the exponential map $P\to K$ (restricted to $P$, as explained above, where it is a homeomorphism). So $h$ is $1$-Lipschitz (and not injective on all of $Q$, although it is a homeomorphism on $P$). Also, $h$ commutes with all the symmetries of $(Q,K)$ because the construction is completely symmetric. In particular, the various copies of $h\colon Q_i\to K_i$ match at the boundaries, so we get a map $h\colon C\to S$ as announced.

What follows is an illustration of $P$ for $d=2$ (the boundary of $P$ is in red, the boundary of $Q$ is, of course, the black square):

This map still needs to be modified a little: let $h\colon Q\to K$ be obtained by composing the nearest point projection $Q\to P$ (which makes sense since $P$ is convex compact) with the exponential map $P\to K$ (restricted to $P$, as explained above, where it is a homeomorphism). So $h$ is $1$-Lipschitz (and not injective on all of $Q$, although it is a homeomorphism on $P$). Also, $h$ commutes with all the symmetries of $(Q,K)$ because the construction is completely symmetric. In particular, the various copies of $h\colon Q_i\to K_i$ match at the boundaries, so we get a map $h\colon C\to S$ as announced.