Let $X$ be a compact Hausdorff space. If $X$ is finite, then there are at most finitely many open sets and so it is second-countable. On the flipside if $X$ is uncountably infinite, it is possible that $X$ not be second-countable. A classic example is $\omega_1+1$ given the order topology (here $\omega_1$ is the first uncountable ordinal).
But what about if $X$ is countably infinite? Glancing at the abstract of this paper, it seems that the existence of a non-second-countable, countably infinite, compact Hausdorff space is plausible if one forbids the use of the axiom of choice.
It turns out I'm perfectly happy using the axiom of choice. So I'm left wondering whether it is true that a countably infinite, compact Hausdorff space must be second-countable in ZFC. All the examples I concoct work but a proof eludes me. Could someone provide a proof, reference or counterexample, it would be much appreciated!