 Solution

True in ZF. Suppose $\mathcal{P}(A)$ were countably infinite. The singleton injection would inject $A$ into the natural numbers. The set $A$ cannot be finite, since its power set would then be finite. Its image is therefore an infinite subset of $\mathbb{N}$, which has a canonical increasing enumeration. This gives a bijection $A\cong \mathbb{N}$ without choice.

But then $\mathcal{P}(A)\cong \mathcal{P}(\mathbb{N})$, contradicting Cantor's theorem that the latter is strictly larger than $\mathbb{N}$. HenceThe proof does not assume that an arbitrary infinite set contains a countably infinite subset; here the given injection into the natural numbers supplies its well-ordering.