Every infinite set has a countably infinite subset

Statement

Assume the axiom of choice. Let $A$ be an infinite set. Then there is an injective function $f \colon \mathbb{N} \to A$, so $A$ has a countably infinite subset $f(\mathbb{N})$.

Remarks

The proof chooses $a_0 \in A$, then $a_1 \in A \setminus \{a_0\}$, and so on. Each set $A \setminus \{a_0, \dots, a_n\}$ is nonempty because $A$ is infinite. The axiom of countable choice is enough, but some choice is needed: in ZF alone, the statement cannot be proved. It follows that $\aleph_0$ is the least infinite cardinal.

Explanations

L1 human

In an infinite set you can always pick one more element. Picking forever gives a sequence of different elements, which is a copy of $\mathbb{N}$ inside the set.