Infinite Set has Countably Infinite Subset

Theorem
Every infinite set has a countably infinite subset.

Intuitive Proof
Let $S$ be an infinite set, and let $a_0 \in S$.

$S$ is infinite, so $\exists a_1 \in S, a_1 \ne a_0$, and $\exists a_2 \in S, a_2 \ne a_0, a_2 \ne a_1$, and so on.

That is, we can continue to pick elements out of $S$, and assign them the labels $a_0, a_1, a_2, \ldots$ and this procedure will never terminate as $S$ is infinite.

Each one of the elements is in one-to-one correspondence with the elements of $\N$, and therefore the set $\left\{{a_0, a_1, a_2, \ldots}\right\} \subseteq S$ is countably infinite.

Formal Proof
The formal proof follows the same steps as the intuitive one. The first (and most important) part of the proof is to construct an injective function $f:\N\to S$.

In each non-empty subset $A\subset S$, let us choose one element $x_A \in A$. Then, we define $f$ as follows. Let $f\left(1\right)=x_S$. Now, if we have already defined $f\left(1\right),\ldots,f\left(n\right)$, we write $A_n=S\setminus \left\{f\left(1\right),\ldots,f\left(n\right)\right\}$. Since $S$ is infinite, $A_n$ is infinite for each $n\in\N$. Now, we put $f\left(n+1\right)=x_{A_n}$. This finishes the definition of $f$.

To show that $f$ is injective, let $m,n\in\N$, say $m<n$. Then $f\left(m\right)\in\left\{f\left(1\right),\ldots,f\left(n-1\right)\right\}$, but $f\left(n\right)\in S\setminus \left\{f\left(1\right),\ldots,f\left(n-1\right)\right\}$, hence $m\neq n$.

So, now that we have an injection from $\N$ to $S$, let $S_0=f\left(\N\right)$, that is, $S_0$ is the image of $f$. Then $S_0$ is countable, because, since $f$ is injective, it is a bijection from $\N$ onto its image. Another way to see that $S_0$ is countable is to write $S_0=\left\{f\left(1\right),f\left(2\right),\ldots\right\}$

Realizing that $S_0\subset S$ completes the proof.

Note
Choosing an element $x_A$ for every $A\subset X$ would require the Axiom of Choice. However, this theorem does not depend on the AoC, because we could have chosen just the countably many $x_{A_n}$, and we could have done this step-by-step, i.e., we could have chosen each $x_{A_n}$ before defining $f\left(n+2\right)$.

Comment
What this in effect shows is that countably infinite sets are the smallest possible infinite sets.