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This proof uses neither the axiom of choice nor a preliminary choice of points in all the given sets. For each nonempty $A_j$, form a disjoint union $X_j=A_j\bigsqcup \{{\infty }_j\}$ with a distinguished added point, and give it the topologyThe original set $A_j$ has the indiscrete subspace topology, and is clopen. Any open cover of $X_j$ either contains $X_j$, or contains both $A_j$ and $\{{\infty }_j\}$, so it has a subcover of at most two members. Thus $X_j$ is compact by a proof requiring no choices. No Hausdorff restriction is imposed in this question, and these generally non-Hausdorff spaces are allowed.

The product $X={\prod }_{j\ge 1}{X}_j$ has the explicitly given all-infinity point, so it is nonempty without choice. By the assumed countable-product compactness it is compact. For each $j$ letThe coordinate projection is continuous and $A_j$ is closed in $X_j$, so $F_j$ is closed. Any finite intersection of the $F_j$ is nonempty: pick an element of each of those finitely many nonempty sets and put ${\infty }_k$ in every other coordinate. Finite choice follows by induction in ordinary set theory without assuming the axiom of choice.

Compactness and the finite intersection property now give ${\bigcap }_{j\ge 1}{F}_j\ne \varnothing$. Choose one point $x$ from this single nonempty set. The function $f(j)=x_j$ satisfies $f(j)\in A_j$ for every $j$, proving the axiom of countable choice.

The implication holds in choice-free set theory: countable product compactness implies countable choice. The printed lower index zero is inconsistent with the stated positive-integer index set; indexing from one merely relabels this argument and changes no conclusion.