Compactness of a family of closed subsets in a compact space. 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. 
 A: 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. 
