TITLE: Existence in ZF of a set with countable power set
QUESTION [12 upvotes]: Is it consistent with ZF for there to exist a set $S$ such that the power set $P(S)$ is countable? If so, what is the weakest form of the axiom of choice needed to prove that no such set exists?

REPLY [4 votes]: Asaf has answered the question. Let me just add that even more is true: Provably in $\mathsf{ZF}$, If $A$ is a set, and $\aleph_0\le|\mathcal P(A)|$, then in fact $|\mathbb R|=2^{\aleph_0}\le|\mathcal P(A)|$. This is a result of Kuratowski. What one proves is that from an injection from $\mathbb N$ into $\mathcal P(A)$, an infinite sequence of pairwise disjoint subsets of $A$ can be obtained (this takes some work), so there is a surjection from $A$ onto $\mathbb N$, and from this we easily get an injection of $\mathcal P(\mathbb N)$ into $\mathcal P(A)$.
It may well be the case that we have an infinite set $A$ such that $\aleph_0\not\le|\mathcal P(A)|$, though we always have $\aleph_0\le|\mathcal P(\mathcal P(A))|$ for $A$ infinite.