As a matter of taste I prefer an argument more elementary than Eric Wofsey’s to show that there are at most $2^\kappa$ compact Hausdorff spaces of cardinality $\kappa$.
Lemma. Let $X$ be a compact Hausdorff space of cardinality $\kappa$; then $\chi(X)\le\kappa$, i.e., each point has a local base of cardinality at most $\kappa$.
Proof. Fix $x\in X$. For each $y\in X\setminus\{x\}$ there are disjoint open sets $U_y$ and $V_y$ such that $x\in U_y$ and $y\in V_y$. Let $\mathscr{U}=\big\{U_y:y\in X\setminus\{x\}\big\}$, and let $$\mathscr{B}=\left\{\bigcap\mathscr{F}:\mathscr{F}\subseteq\mathscr{U}\text{ and }\mathscr{F}\text{ is finite}\right\}\;.$$ Let $U$ be an open nbhd of $x$; $\{V_y:y\in X\setminus U\}$ is an open cover of the compact set $X\setminus U$, so there is a finite $F\subseteq X\setminus U$ such that $\{V_y:y\in F\}$ covers $U$. It follows that $\bigcap_{y\in F}U_y\subseteq U$ and hence that $\mathscr{B}$ is a local base at $x$. $\dashv$
It’s an immediate corollary that $w(X)\le|X|\cdot\chi(X)\le\kappa$, i.e., that $X$ has a base of cardinality $\kappa$.
$X$ is Tikhonov of weight at most $\kappa$, so $X$ can be embedded in the cube $[0,1]^\kappa$ of weight $\kappa$. $[0,1]^\kappa$ has at most (in fact exactly) $2^\kappa$ open subsets, so it has at most $2^\kappa$ closed subsets. $X$ embeds as one of these subsets, so there are at most $2^\kappa$ possibilities for $X$.
Lemma: Let $(X,d)$ be a metric space. For each $n \in \{1,2,3,\ldots\}$ there is (by Zorn's lemma) a maximal (by inclusion) set $D_n$ such that for all $x,y \in D_n$ with $x \neq y$, we have $d(x,y) \ge \frac{1}{n}$. If all $D_n$ are at most countable, then $X$ is separable.
Proof: We'll show that $D = \cup_n D_n$ is dense (and is countable when all $D_n$ are), and so pick any $x \in X$ and any $r>0$ and we'll show that $B(x,r) \cap D \neq \emptyset$. Suppose not, then find $m$ with $\frac{1}{m} < r$. Then from $B(x,r) \cap D = \emptyset$ we know that
$d(x, y) \ge \frac{1}{m}$ for all $y \in D_m$ (or else $y \in D \cap B(x,r)$), and then $D_m \cup \{x\}$ would contradict the maximality of $D_m$. So the intersection with $D_m$ and hence $D$ is non-empty, and so $D$ is dense.
(The above is the heart of the proof that a ccc metric space is separable, as $X$ ccc implies that all $D_m$ are countable, as the $B(x,\frac{1}{m})$, $x \in D_m$ are a pairwise disjoint open family of non-empty sets.)
Now suppose that $(X,d)$ is countably compact. Suppose for a contradiction that $X$ is not separable. Then for some $m \ge 1$ we have that $D_m$ (as in the lemma) is uncountable.
Fix such an $m$.
Now $D_m$ is discrete (clear as $B(x,\frac{1}{m}) \cap D_m = \{x\}$ for each $x \in D_m$) and closed: suppose that $y \in X\setminus D_m$ is in $\overline{D_m}$, then $B(y, \frac{1}{2m})$ contains infinitely many points of $D_m$ and for any $2$ of them, say $x_1, x_2 \in D_m$, we'd have $d(x_1, x_2) \le d(x_1, y) + d(y,x_2) < \frac{1}{m}$ contradiction (as points in $D_m$ are at least $\frac{1}{m}$ apart). So $D_m$ is closed and discrete (as an aside: being uncountable this would already contradict Lindelöfness of $X$; this shows that a Lindelöf metric space is separable, e.g.), but we want to contradict countable compactness, which is easy too:
Choose $A \subseteq D_m$ countably infinite (so that $A$ is closed in $X$, as all subsets of $D_m$ are), and define a countable open cover $\mathcal{U} = \{B(p, \frac{1}{m}): p \in A\} \cup \{X\setminus A\}$ of $X$ that has no finite subcover: we need every $B(p,\frac{1}{m})$ to cover $p$ for all $p \in A$.
This contradiction then shows $X$ is separable (and thus has a countable base ,is Lindelöf etc. finishing the compactness).
The crucial fact is that all of the following are equivalent for a metric space:
- $X$ has a countable base.
- $X$ is separable.
- All discrete subspaces of $X$ are at most countable.
- All closed and discrete subspaces of $X$ are at most countable.
- $X$ is ccc.
- $X$ is Lindelöf.
In the above I essentially did "not (2) implies not (4)" implicitly. The fun is that countable compactness implies (4) easily (such closed discrete subspaces are even finite) and thus we get Lindelöfness "for free". Also implicit in the above proof is that every countably compact space is limit point compact.
Best Answer
HINT:
There is no set of all compact metric spaces, because every singleton is a compact metric space, and the collection of all singletons is not a set.
However, note that every compact metric space is homeomorphic to a closed subspace of the Hilbert cube. So if you are only interested in equivalence classes of compact metric spaces, it suffices to consider subspaces of the Hilbert cube.