= Solution
\b[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 $\mathbb N$. Hence
$$
\boxed{2^\kappa\ne\aleph_0\quad\text{for every cardinal }\kappa.}
$$
The 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>.
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