 Double-power-set countability bound without choice

For each natural number $n$, the family $[A{]}^n$ of all $n$-element subsets is nonempty when $A$ is not finite. These families are pairwise distinct elements of $\mathcal{P}(\mathcal{P}(A))$, giving an explicit injection of the natural numbers. No representative from each family is selected. A power set cannot be countably infinite in ZF: if $\mathcal{P}(B)$ were countably infinite, singleton injection would make $B$ an infinite subset of a countable set and hence countably infinite; Cantor's theorem would then contradict countability of its power set. Apply this to $B=\mathcal{P}(A)$ to make the displayed lower bound strict.