Completeness of the Hausdorff distance. THEOREM
Let us define $$F(X) := \{ A \subseteq X | A \neq \emptyset, A - \text{closed and bounded} \} $$
Then if $(X, \rho ) $ is complete, then $(F(X), d_h )$ is complete.

Are there any interesting proofs of this theorem other than the ones with long computations with $\epsilon$? I have been searching the internet for a while, but could not find anything.
HINT:

Step 1. Reduction to a simpler case. Assume that we have $A_n$ bounded and closed such that $d(A_m, A_n) \to 0$. Consider $A'_n \colon = \overline{\cup_{m \ge n} A_m}$.  Check that $d(A_n, A'_n) \to 0$ ( simple). Now we have $(A'_n)$ decreasing and Cauchy.

Step2: Proof in the simpler case : if $A_n$ is a decreasing sequence of nonvoid, closed and bounded subsets of $X$ such that $d(A_n, A_m) \underset{m, n \to \infty}{\longrightarrow} 0$, then $A\colon =\cap A_n$ is nonvoid and $A_n \to A$.

A bit of $\epsilon$ work : take $\epsilon &gt; 0$ and consider $(n_k)$ a subsequence of the naturals so that $d(A_m, A_n) \le \epsilon/2^k$ for $m,n \ge n_k$. Every point in $A_{n_1}$ is at distance at most $\epsilon$ from a point in $A$. Indeed, take $x_1$ in $A_{n_1}$ arbitrary, then $x_2 \in A_{n_2}$ , $d(x_2, x_1) \le \epsilon/2$, then $x_3 \in A_{n_3}$, $d(x_3, x_2) \le \epsilon/2^2$, and so on. The sequence $x_n$ converges to a point in $x$ in $A$ and we have $d(x, x_1) \le \epsilon/2 + \epsilon/2^2 + \cdots = \epsilon$. 

Basically we show that $A_n$ converges to $\lim \sup A_n$ ( limsup in the lattice of closed subsets).
Hint: You know that the Hausdorff distance is equivalently defined by $d_h(A, B)=\max\{\sup\limits_{a\in A}\rho(a, B),\sup\limits_{b\in B}\rho(b, A)\}.$ Pick a Cauchy sequence $A_n\subset F(X)$ and show that it converges with this metric.