Let $M$ be a metric space, and
$$2^M=\{A\subset M\mid A\textrm{ is compact and not empty}\}.$$
Given $A\in2^M$ and $\epsilon>0$, define
$$N(\epsilon,A)=\{x\in M\mid d(x,a)<\epsilon\textrm{ for some }a\in A\},$$
and for any $A,B\in 2^M$ let
$$H(A,B)=\inf\{\epsilon>0\mid A\subset N(\epsilon,B)\textrm{ and }B\subset N(\epsilon,A)\}.$$
Show that if $(M,d)$ is complete, then $(2^M,H)$ is also complete.
What I've tried
Let $\{A_i\}_{i=1}^\infty\subset2^M$ be a Cauchy sequence. Then, given $\epsilon>0,$ $\exists N$ such that $n,m\geq N\Rightarrow A_n\subset N(\epsilon,A_m)$ and viceversa(*). Define
$$S=\{\{a_i\}_{i=1}^\infty\subset M|a_i\in A_i\textrm{ and }\{a_i\}_{i=1}^\infty \textrm{ is a Cauchy sequence}\}.$$
By (*), $S$ is not empty. Since $M$ is complete, every $\{a_i\}_{i=1}^\infty\in S$ converge to some point in $M$. So, let $L=\{\lim_{i\to\infty} a_i|\{a_i\}_{i=1}^\infty\in S\}$.
That's pretty much what I've got. I believe that $A_i\to L$, but I'm clueless about how to prove that. I don't even know how to prove that $L\in 2^M$.
An idea is to use the concepts of $\lim\inf$ and $\lim\sup$: Given any sequence $\{A_i\}_{i^1}^\infty\subset 2^M$, define
$$\lim\inf A_i=\{x\in M|\textrm{for each open set }U\textrm{ such that }x\in U, U\cap A_i\neq\emptyset\textrm{ for all but finitely many }i\textrm{'s}\},$$
$$\lim\sup A_i=\{x\in M|\textrm{for each open set }U\textrm{ such that }x\in U, U\cap A_i\neq\emptyset\textrm{ for infinitely many }i\textrm{'s}\}.$$
It can be shown that for arbitrary $M$, $A_i\to A\Rightarrow \lim\inf A_i=\lim\sup A_i=A$. However, the only proof I know for the converse is for compact $M$. So that seems like a dead end.
These very detailed notes (a student project) has all the details for this result, including all definitions and some examples. The limit for a Cauchy sequence $(A_n)$ is the set of all limits of sequences $(a_n)$ where $a_n \in A_n$, so basically your set $L$, as Cauchy is equivalent to convergent in $(X,d)$. As a bonus you get a proof there that $X$ compact implies $2^X$ compact as well.