Gamelin and Greene's "Introduction to Topology", Second Edition, Dover, pg25, Exercise 8, states that if $X$ is a compact space, then the family of non-empty, closed subsets, $C$, with the metric $$\rho(E,F) = \max \left( \sup_{x\in E} d(x,F) ,\ \sup_{y\in F} d(y,E) \right),$$ is compact. The $d(u,v)$ is the metric for the given compact space $X$. $E$ and $F$ are members of the family $C$.
On page 200, the authors supply a proof which I find too terse to follow, though I must admit - with some embarrassment - I have spent much time. On the internet, I have found proofs of the well known theorem about a subset of a compact set, but not another proof of the above. I am seeking a more expansive proof of Gamelin and Greene's exercise.
This proof is definitely longer than that in the book. Whether or not it is clearer is up to you to judge.
We will show that $(\mathcal C, \rho)$ is complete and totally bounded.
Claim one $(\mathcal C, \rho)$ is complete: Let $\{E_n\}$ be a Cauchy sequence in $(\mathcal C, \rho)$. Then define $$E = \{ x\in X: \text{there is }x_{n_k} \in E_{n_k} \text{ so that }x_{n_k} \to x\}.$$ First of all, $E$ is nonempty: Pick any sequence $x_n \in E_n$. As $X$ is compact, a subsequence of $\{x_{n_k}\}$ of $\{x_n\}$ converges to some point $x\in X$. Then by definition $x\in E$.
Secondly, $E$ is closed: Let $x_1, x_2, \cdots $ be a sequence in $E$ so that $x_n \to y$. Then for each $x_k$, by definition of $E$, there is $x_{n_k} \in E_{n_k}$ so that $d(x_k , x_{n_k})<\frac{1}{k}$. As $x_n \to y$, then so is $x_{n_k} \to y$. Thus $y\in E$ and so $E$ is closed.
Now as $X$ is compact and $E$ is closed, $E$ is also compact. Together with the fact that $E$ is nonempty, $E\in \mathcal C$.
Next we show $E_n \to E$ in $(\mathcal C, \rho)$. Let $\epsilon >0$. Then as $\{E_n\}$ is Cauchy, there is $N$ so that $\rho(E_n, E_m) < \epsilon/3$ for all $m, n\ge N$. Let $x\in E$. Then by definition of $E$, there is $E_n$ (where $n$ depends on $x$ and we choose $n \ge N$) and $x_n\in E_n$ so that $$d(x, x_n ) < \epsilon/3. $$ Then for all $y\in E\cap B(x, \epsilon/3)$, we have $$ d(y, x_n ) < d(y, x) + d(x_n, x) < \frac{2\epsilon}{3}.$$ This implies $$d(y, E_n) <\frac{2\epsilon}{3}$$ for all $y\in E \cap B(x, \epsilon /3)$. Now as $n\ge N$, for all $m\ge n$ we have $$ d(y, E_m) \le d(y, En) + \rho (E_n, E_m) < \epsilon.$$ Then as $E$ is compact, there is $x_1, x_2, \cdots, x_l \in E$ so that $$\{ B(x_i, \epsilon/3)| i = 1, 2, \cdots, l\}$$ covers $E$. Let $N_1 = \max_{i} \{ n_{x_i}\}$. Then for all $m\ge N_1$, we have $$\tag{1} d(y, E_m) < \epsilon,\ \ \ \ \ \forall y\in E.$$ Now we claim:
Claim 2: there is $N_2$ so that $$ \sup_{x\in E_m}d(x, E) < \epsilon \ \ \ \ \ \forall m\ge N_2.$$ If not, there is a subsequnce $\{n_k\}$ so that $$ \sup_{x\in E_{n_k}}d(x, E) \ge \epsilon . $$ In particular, there is $x_{n_k} \in E_{n_k}$ so that $$\tag{2} d(x_{n_k}, E) \ge \epsilon/2. $$ But as $X$ is compact, by passing to a subsequence we have $x_{n_k} \to x$. By definition of $E$, $x\in E$. But this contradicts $(2)$. Thus claim 2 is shown.
Now as choose $N_\epsilon = \max\{N_1, N_2\}$. Then for all $m\ge N$, we have $$ \rho(E, E_m) \le \epsilon.$$ Thus $E_n \to E$ in $(\mathcal C, \rho)$. This proves claim one that $(\mathcal C, \rho)$ is complete.
Claim three $(\mathcal C, \rho)$ is totally bounded. Let $\epsilon >0$ Then as $X$ is compact, there is $N \in \mathbb N$ and $x_1, \cdots, x_N \in X$ so that $$ X = \bigcup_{i=1}^N \overline{B(x_i, \epsilon)}.$$
Now the power set $\mathcal P_N$ of $\{x_1, \cdots, x_N\}$ (excluding the emptyset) can be considered a finite subset of $\mathcal C$ as all finite points are compact.
Claim four: We have $$(\mathcal C, \rho) = \bigcup_{p\in \mathbb P_N} \overline{B_\rho (p, \epsilon)}.$$
To see this, let $E\in \mathcal C$ and $I \subset P_N$ so that$ x_i\in I$ if and only if $\overline{B(x_i, \epsilon)} \cap E \neq \emptyset$. Then
$$\begin{split} \rho(I, E) &= \max\left( \sup_{x_i\in I} d(x_i, E), \sup_{x\in E} d(x, I)\right) \\ &\le \max ( \epsilon, \epsilon) = \epsilon. \end{split}$$
Thus the Claim four is shown.
To sum up, $(\mathcal C, \rho)$ is complete and totally bounded, so $(\mathcal C, \rho)$ is compact.