I am trying to understand a particular proof that a subset of a countable set is countable. I'm defining "countable" as finite or countably infinite. The exact statement is:
If $S$ is countable and $T \subset S$, then $T$ is countably infinite.
If $S$ is finite, then $T \subset S$ is finite, hence countable, so suppose that $S$ is countably infinite.
The step I am struggling to fully grasp is this: "without loss of generality, we can assume $S = \mathbb{N}$ and $T \subset \mathbb{N}$."
After this assumption is made, the proof is clear to me by the well-ordering principle, but I am trying to understand why exactly I can make this simplification. As $S$ is countably infinite, I can find a bijection $f: \mathbb{N} \to S$ and another $f^{-1} : S \to \mathbb{N}$. If $T \subset S$, then $f^{-1} (T) \subset f^{-1} (S) \subset \mathbb{N}$.
I suppose the argument is that, because $f^{-1}$ is bijective, $|f^{-1} (T)| = |T|$, so if I prove that $f^{-1}(T)$ is countable, I prove that $T$ is countable. I don't know if this is the same as saying "$S = \mathbb{N}$," but I think this is roughly on the correct track.
Could someone help me fill in the missing details?
Assume you've proved the result in the special case where $S = \mathbf{N}$.
Now assume instead that $S$ is an arbitrary countably infinite set, and $T$ is a subset of $S$. Since $S$ is countably infinite, there exists a bijection $f$ of $S$ onto $\mathbf{N}$.
The set $f(T)$ is a subset of $\mathbf{N}$, hence by the special case we're assuming to have been proved, $f(T)$ is countable. But because $f$ defines a bijection of $T$ onto $f(T)$, the set $T$ must be countable. This proves the statement in general.