Prove that every infinite set has a countable subset.

I just need help trying to create a proof that shows that an infinite set has a countable subset. Is it as simple as taking arbitrary values of the finite set and listing them in their own subset?

Definition: The statement that a set $S$ is infinite means that if $N$ is a natural number then $S$ contains $N$ distinct elements.

[Note: If an infinite set is defined in this way, then it automatically follows that an infinite set minus a finite set is infinite.]

Suppose $S$ is infinite.

Since $1$ is a natural number, $S$ contains an element $x_1$.

Since $2$ is a natural number, $S$ contains an element $x_2$ distinct from $x_1$.

So there is a two-element subset $U_2=\{x_1,x_2\}$ of distinct elements of $S$.

For each $N\in\mathbb{N}$, $S$ contains an element distinct from each element in $U_N=\{x_1,x_2,\cdots,x_n\}$, so define $U_{N+1}=U_N\cup\{x_{N+1}\}$ where $x_{N+1}$ is an element of $S$ distinct from each element of $U_N$.

Let $$U=\bigcup_{N\in\mathbb{N}}U_N$$

Then $U$ is a countable subset of $S$.

ADDENDUM There is an issue which I glossed over when making this argument.

For each $N$ we know that there is a subset $V$ of $S$ containing $N+1$ elements of $S$. So $V$ contains an element, call it $x_{N+1}$, which is distinct from each element of $U_N=\left\{x_n,x_2,\cdots,x_N\right\}$. Let $U_{N+1}=\left\{x_n,x_2,\cdots,x_N,x_{N+1}\right\}$.

Looking over the comments between John Wayland Bales and fleablood, I thought I would take a stab at this 'iconoclast approach', working in the realm of intuitive set theory.

Definition: A set is $S$ finite if there exist a natural number $n \gt 0$ such that every mapping $f: \{1, 2, \dots, n\} \to S$, the mapping $f$ is not an injection.

For a finite set $S$ we can consider the smallest $n_0$ where there are no injections. We then can say that $S$ has $n_0 - 1$ elements.

Proposition 1: If a finite set $S$ has $m$ elements then $S$ is equinumerous with $\{1,2,\dots,m\}$.
Proof
Let $g$ be an injection of $\{1,2,\dots,m\}$ into $S$. This mapping is necessarily a surjection. Suppose, to get a contradiction, that an element $s_0 \in S$ is not in the range of $g$. Then we can extend $g$ to another injective function by defining $g(m+1) = s_0$, but this is absurd. $\quad \blacksquare$

A set $S$ is infinite if it is not finite. This means that for every $n \gt 0$ there exist an injective mapping from $\{1,2,\dots,n\}$ into $S$.

Considering Dan Christensen's answer to the OP's question, we introduce the following axiom:

A set $S$ is infinite if and only there exist a function $f: S \to S$ that is an injection but is not a surjection.