Cantor's theorem

There is no surjection from a set $X$ onto its power set $\mathcal{P}(X)$. Consequently $\mathcal{P}(X)$ has strictly greater cardinality than $X$.

For any candidate function $f:X\to \mathcal{P}(X)$, form $D=\{x\in X:x\notin f(x)\}$. This is a subset of $X$ by set comprehension. If $f(d)=D$, then $d\in D$ holds exactly when $d\notin D$, a contradiction. Thus $D$ is missing from the image of $f$. On the other hand, $x\mapsto \{x\}$ is an injection from $X$ into its power set, establishing the strict comparison of cardinalities. For $X=\mathbb{N}$, the omitted selection is the subset version of Cantor's diagonal argument.

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.