TITLE: How to show that Hausdorff distance is a metric on the set of all compact non-empty subsets of a Polish space?
QUESTION [5 upvotes]: For each perfect Polish space $X$, let $H[X]$ be the set of all compact non-empty
  subsets of $X$. If $x ∈ X$ and $A ∈ H[X]$, put
  $$d(x,A) = \inf \{d(x, y) : y ∈ A\}$$
  where on the right $d$ is the distance function on $X$. The Hausdorff distance between
  two compact sets is defined by
  $$d(A,B) = \max \{ \sup \{d(x,B) : x ∈ A\},
\sup\{d(y,A) : y ∈ B\}\}$$
  Prove that this is a metric on $H[X]$.

This is an exercise on page 13, Descriptive Set Theory, Yiannis N. moschovakis(2009). I got stuck on how to show $d(A,B)+d(B,C) \ge  d(A,C)$.

REPLY [12 votes]: We have
$$d(a,C)\le d(a,b)+d(b,C)\le d(a,b)+d(B,C) $$
for all $b\in B$, so, taking $\inf_{b\in B}$, we have 
$d(a,C)\le d(a,B)+d(B,C)$ 
which is $\le d(A,B)+d(B,C)$
then take $\sup_{a\in A}$, and similarly, we can show that $d(A,c)\le d(A,B)+d(B,C)$ as well.