Compact Hausdorff spaces are metrizable if and only if second-countable

1.1. Let \((X, d)\) be a compact metric space.

1.2. For each positive integer \(m \in \mathbb{N}\), the collection of open balls \(\{B(x, 1/m) : x \in X\}\) is an open cover of \(X\).

1.3. By compactness of \(X\), there exists a finite set of points \(D_m = \{x_{m,1}, x_{m,2}, \dots, x_{m,k_m}\} \subset X\) such that\[X = \bigcup_{i=1}^{k_m} B(x_{m,i}, 1/m).\]

1.4. Define \(\mathcal{B} = \bigcup_{m=1}^\infty \{B(x_{m,i}, 1/m) : 1 \le i \le k_m\}\). As a countable union of finite sets, \(\mathcal{B}\) is countable.

1.5. We show \(\mathcal{B}\) is a basis: let \(U \subseteq X\) be open and \(x \in U\). Choose \(r > 0\) such that \(B(x, r) \subseteq U\), and choose \(m \in \mathbb{N}\) with \(1/m < r/2\).

1.6. Since \(D_m\) covers \(X\), there is some \(x_{m,i} \in D_m\) with \(x \in B(x_{m,i}, 1/m)\).

1.7. For any \(y \in B(x_{m,i}, 1/m)\), the triangle inequality gives\[d(y, x) \le d(y, x_{m,i}) + d(x_{m,i}, x) < \frac{1}{m} + \frac{1}{m} = \frac{2}{m} < r,\]so \(x \in B(x_{m,i}, 1/m) \subseteq B(x, r) \subseteq U\).

1.8. Thus \(\mathcal{B}\) is a countable basis for the topology of \(X\), so \(X\) is second-countable.

2.1. Every compact Hausdorff space is normal (\(T_4\)).Proof of normality: For any closed set \(F \subset X\) and point \(y \notin F\), since \(X\) is Hausdorff, each \(x \in F\) has disjoint open neighborhoods \(U_x \ni x\) and \(V_x \ni y\). By compactness of \(F\), finitely many \(U_{x_i}\) cover \(F\), and \(\bigcap V_{x_i}\) is an open neighborhood of \(y\) disjoint from \(\bigcup U_{x_i}\). Repeating this argument for two disjoint closed sets separates them by disjoint open neighborhoods.

2.2. Every normal \(T_1\) space is regular (\(T_3\)).

2.3. By the Urysohn Metrization Theorem, every second-countable regular space is metrizable.

2.4. Therefore \(X\) is metrizable.