 Solution

True in ZF. Let $A$ be a set that is not finite. For each natural number $n$, formEach $E_n$ is a subset of $\mathcal{P}(A)$ and is nonempty: a nonfinite set has an $n$-element subset for every finite $n$, by finite induction. The sets $E_n$ are pairwise disjoint, hence pairwise distinct. Consequentlyis an injection from $\mathbb{N}$ into $\mathcal{P}(\mathcal{P}(A))$. This is a double-power-set countability bound without choice.

Part (d), applied to the set $\mathcal{P}(A)$, says that its power set is not countably infinite. Combining that result with the displayed injection givesThere is no hidden countable choice: the proof uses the entire definable family of $n$-element subsets, rather than choosing one subset for each $n$.