Are countably infinite, compact, Hausdorff spaces necessarily second countable? 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! 
 A: In fact, thanks to Sierpinski-Mazurkiewicz theorem, one precisely knows all Hausdorff countable compact topological spaces; namely, any such space is homeomorphic to $\omega^{\alpha} \cdot n +1$ for some countable ordinal $\alpha$ and some integer $n \geq 1$. See for example my answer here. Now, it is not difficult to verify that $\omega^{\alpha} \cdot n +1$ is second-countable.
A: Let $X$ be a countably infinite compact Hausdorff space. I will show (in ZFC) that $X$ is second-countable.
Since $X$ is countable, it suffices to show that each point $x\in X$ has a countable neighborhood base. All these neighborhood bases taken together will be a countable base for the topology of $X$.
So fix $x\in X$, and for each other point $y\in X$, fix disjoint open sets $U_y$ and $V_y$ such that $x\in U_y$ and $y\in V_y$. let $\mathcal{U}$ be the collection of all open sets of the form $\bigcap_{i = 1}^n U_{y_i}$ for $y_1,\dots,y_n\in X$. $\mathcal{U}$ is countable, and I claim it is a neighborhood base for $x$.
Indeed, let $O$ be an open set containing $x$, and let $C = X\setminus O$. Then $C$ is closed, hence compact. The collection of open sets $\{V_y\mid y\in C\}$ covers $C$, so it can be refined to a finite subcover, $\{V_{y_1},\dots,V_{y_n}\}$. Since $C \subseteq \bigcup_{i = 1}^n V_{y_i}$, the set $\bigcap_{i = 1}^n U_{y_i}$ in our proposed neighborhood base is a subset of $O$, as was to be shown.
A: They mention this in the paper. Essentially this is Theorems 2.2+2.3, and the obvious implication that the axiom of choice implies what they call $\bf CAC(\Bbb R)$.
