Triangle inequality of a nasty metric $\rho(A,B)=\max\left\{\sup_{x\in A}\inf_{y\in B}|x-y|,\sup_{y\in B}\inf_{x\in A}|x-y|\right\}$ This was a qual problem in Winter 2012 at my university. I feel comfortable verifying everything but the triangle inequality.

Let $\Omega$ denote the set of all nonempty closed subsets of $[0,1]$ and let $\rho\colon\Omega\times\Omega\to[0,1]$ be defined by
  $$
\rho(A,B)=\max\left\{\sup_{x\in A}\inf_{y\in B}|x-y|,\sup_{y\in B}\inf_{x\in A}|x-y|\right\}$$
  Show that $(\Omega,\rho)$ is a metric space.

Intuitively, I think of $\sup_{x\in A}\inf_{y\in B}|x-y|$ as the closest distance from each $x\in A$ to the set $B$, and then taking the largest of those.
Is there a clever way to see $\rho$ respects the triangle inequality?

This page at planetmath.org has a very brief sketch of the proof of the triangle inequality; you might like to read it and see whether you can fill in the details. This PDF has a complete proof in case you get stuck.
