Proposition   Every compact metric space is second countable.

Proof. Let $(X,d)$ be a compact metric space, and for each $n\in\mathbb{Z}^+$ define $\mathcal{A}_n=\{B(x,1/n):x\in X\}$ , where $B(x,1/n)$ denotes the open ball centered about $x$ of radius $1/n$ . Each such collection is an open cover of the compact space $X$ , so for each $n\in\mathbb{Z}^+$ there exists a finite collection $\mathcal{B}_n\subseteq\mathcal{A}_n$ that covers $X$ . Put $\mathcal{B}=\bigcup_{n=1}^\infty\mathcal{B}_n$ . Being a countable union of finite sets, it follows that $\mathcal{B}$ is countable; we assert that it forms a basis for the metric topology on $X$ . The first property of a basis is satisfied trivially, as each set $\mathcal{B}_n$ is an open cover of $X$ . For the second property, let $x,x_1,x_2\in X$ , $n_1,n_2\in\mathbb{Z}^+$ , and suppose $x\in B(x_1,1/n_1)\cap B(x_2,1/n_2)$ . Because the sets $B(x_1,1/n_1)$ and $B(x_2,1/n_2)$ are open in the metric topology on $X$ , their intersection is also open, so there exists $\epsilon>0$ such that $B(x,\epsilon)\subseteq B(x_1,1/n_1)\cap B(x_2,1/n_2)$ . Select $N\in\mathbb{Z}^+$ such that $1/N<\epsilon$ . There must exist $x_3\in X$ such that $x\in B(x_3,1/2N)$ (since $\mathcal{B}_{2N}$ is an open cover of $X$ ). To see that $B(x_3,1/2N)\subseteq B(x_1,1/n_1)\cap B(x_2,1/n_2)$ , let $y\in B(x_3,1/2N)$ . Then we have \begin{equation} d(x,y)\leq d(x,x_3)+d(x_3,y)<\dfrac{1}{2N}+\dfrac{1}{2N}=\dfrac{1}{N}<\epsilon\text{,} \end{equation}so that $y\in B(x,\epsilon)$ , from which it follows that $y\in B(x_1,1/n_1)\cap B(x_2,1/n_2)$ , hence that $B(x_3,1/2N)\subseteq B(x_1,1/n_1)\cap B(x_2,1/n_2)$ . Thus the countable collection $\mathcal{B}$ forms a basis for a topology on $X$ ; the verification that the topology induced by $\mathcal{B}$ is in fact the metric topology follows by an argument similar to that used to verify the second property of a basis, and completes the proof that $X$ is second countable.