Infinite Set has Countably Infinite Subset/Proof 1

Theorem
Every infinite set has a countably infinite subset.

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

Suppose there were an injection $\psi: \N \hookrightarrow S$.

Let $\operatorname{Im} \left({\psi}\right) = T$ be the image of $\psi$.

From Injection to Image is Bijection it follows that $\psi^{-1}: T \to \N$ is a bijection.

Thus by definition, $T$ is countably infinite subset of $S$, contrary to hypothesis.

So there could be no such injection.

From Between Two Sets Exists Injection or Surjection, it follows that there is function $\phi: \N \to S$ which is surjective but not injective.

Thus from Surjection iff Right Inverse, $\phi$ has a right inverse $\phi^{-1}: S \to \N$ such that $\phi \circ \phi^{-1} = I_S$.

From Right Inverse Mapping is Injection, it follows that $\phi^{-1}$ is injective.

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.

Hence the result from proof by contradiction.