 Countable product compactness implies countable choice

This concerns arbitrary compact spaces, with no Hausdorff restriction. For nonempty sets $A_j$, adjoin a distinguished point ${\infty }_j$ and use the four-open-set topology $\{\varnothing ,A_j,\{{\infty }_j\},A_j\cup \{{\infty }_j\}\}$. Each resulting space is compact without choice. The product has the canonical all-infinity point. The closed coordinate conditions $x_j\in A_j$ have the finite intersection property, since finite choice is provable without the axiom of choice. If the countable product is compact, their intersection is nonempty, and any point in it supplies a choice function.