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 $f\left(m\right) \neq f\left(n\right)$.

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 each $A\subset X$ requires the Axiom of Choice. Although this is intuitively an "obvious" theorem, it is a theorem about cardinalities. The theory of cardinal numbers depends strongly on the axiom of choice; without AoC one can't guarantee "elementary" and "intuitive" results, like if $f:X\rightarrow Y$ is surjective, then $\exists g:Y\rightarrow X$ injective. It is not true that only "complex" or "strange" results (like the Banach-Tarski Paradox, or the existence of a basis for $\R$ as a vector space over $\Q$) depend on the AoC.

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

Proof
Assume that $S$ is an infinite set which has no countably infinite subset.

Suppose there were an injection from $\N$ to $S$.

Then its image would be a countably infinite subset of $S$.

So there could be no such injection.

Thus there is function $\phi: \N \to S$ which is surjective but not injective.

From Surjection iff Right Inverse there exists an injection $\phi^{-1}: S \to \N$ such that $\phi \circ \phi^{-1} = I_S$.

But from Injection from Infinite to Countably Infinite Set it follows that $S$ is countably infinite.

So from Subset of Itself $S$ has a subset $S$ which is countably infinite.

The proof follows by contradiction.