I recently stumbled across a property of compact Hausdorff spaces which is supposedly well-known, namely:
If $X$ is a compact Hausdorff space, then $X$ is second countable if and only if $C(X)$ is separable.
Now, I was not able to find a proof of this anywhere that does not rely on showing equivalences to metrizability, and was wondering if there is a more elegant way of showing this fact.
If $C(X)$ is separable, take a countable dense set $D\subseteq C(X)$. Then the open sets of the form $$ U=f^{-1}(0,1), $$ for $f$ in $D$, will form a countable basis for the topology of $X$.
Conversely, if $X$ is second countable, take a countable basis of open sets $\scr U$ and, for every pair $(U_1, U_2)$ of open sets in $\scr U$, such that $\overline U_1 \subseteq U_2$, chose an (Urysohn) function $$ f=f_{U_1, U_2}:X\to {\mathbb R} $$ such that $f\equiv 1$ on $U_1$, and $f\equiv 0$ on $X\setminus U_2$. Then the rational subalgebra generated by the $f_{U_1, U_2}$ is countable and dense in $C(X)$ (by Stone-Weiestrass), so $C(X)$ is separable.
EDIT: Upon request, let me prove the first claim above. Actually let me instead prove that the sets of the form $$ U=f^{-1}(1,+\infty ), $$ for $f$ in $D$, form a basis for the topology of $X$. This would in fact be a better claim in my answer and it also clearly implies that $X$ is seconnd countable. If desired, one may then easily use this fact to prove the claim, as it appears above.
Let $V\subseteq X$ be any open set and let $p$ be a point in $V$. It is then enough to find some $f$ in $D$ such that $$ p\in f^{-1}(1, +\infty )\subseteq V.\tag{*} $$ In order to do so, use Urysohn to find a continuous function $g:X\to [0,2]$, such that $g(p)=2$, and $g=0$ off $V$.
Since $D$ is dense in $C(X)$, we may pick some $f$ in $D$ such that $\|g-f\|<1$.
It follows that $f(p)>1$, so $p\in f^{-1}(1,+\infty )$.
Also $|f(x)|≤1$ on $X\setminus V$, so $$ X\setminus V\subseteq f^{-1}(-\infty , 1]. $$ By taking complements (and reversing the inclusion sign), we get $$ f^{-1}(1, \infty )\subseteq V, $$ proving (*).