Why does the (topology given by) Hausdorff metric depend only on the topology? If I have a compact metric space $(X,d)$, I can define the Hausdorff metric on the set $K(X)$ of all non-empty compact (equivalently, closed) subsets of $X$ as $$d_H(A,B) = \max ( \sup_{x \in A} \inf_{y \in B} d(x,y),  \sup_{y \in A} \inf_{x \in B} d(x,y))$$
Now, I'm told, "the topology on $K(X)$ depends only on the topology of $X$, as any two metrics on $X$ are equivalent". Should I interpret the any two metrics are equivalent as "they generate the same topology"? Then, the implication of the Hausdorff metrics being equivalent is not clear at all.
On the other hand, if I interpret two metrics being equivalent as $c d_1<d_2 < Cd_1$, then the Hausdorff metrics producing the same topology seems clear, but the fact that any two metrics on $X$ are equivalent is suspicious. Does the compactness of $X$ force this?
Even if I know one of the interpretation to be correct, I can then try and prove the statement. In fact, I would prefer that, over a complete proof of the statement. 
As an aside, is there a standard notation for my $K(X)$? The book I'm reading uses $2^X$, although I suspect due to typographical limitations.
 A: Your first interpretation is the correct one.
The topology on $K(X)$ induced by the Hausdorff metric has a description as a hit-and-miss topology. That is, for a nonempty open set $U \subseteq X$ put
$$
U^+ = \{C \in K(X) \mid C \cap U \neq \emptyset\}
$$
and
$$
U^- = \{C \in K(X) \mid C \subseteq U\}.
$$
The set $U^+$ contains the compact sets that meet $U$ and the set $U^-$ contains the compact sets that miss $X \setminus U$.
The sets $U^+$ and $U^-$ where $U$ runs through the non-empty open sets of $X$ form a subbasis for the so-called Vietoris topology on $K(X)$. The sets $[U; V_1,\dots,V_n] := U^- \cap V_{1}^+ \cap \cdots \cap V_{n}^+$ form a convenient basis for the Vietoris topology.
The point is that one can prove that the Hausdorff metric induces the Vietoris topology on $K(X)$.
It may be helpful to prove at some point that for a countable dense subset $D$ of $X$ the collection of non-empty finite subsets of $D$ forms a countable dense subset of $K(X)$.
A nice and detailed exposition of these ideas can be found e.g. in Srivastava, A course on Borel sets, section 2.6, see Spaces of Compact sets, pages 66ff.
Added: The second interpretation is too strong. It is not very hard to show that a fat Cantor set and the usual ternary Cantor set are homeomorphic, but not bi-Lipschitz homeomorphic.
