Infinite if Injection from Natural Numbers

Theorem
Let $$S$$ be a set.

Let there exist an injection $$\phi: \N \to S$$ from the natural numbers to $$S$$.

Then $$S$$ is infinite.

Proof
Suppose, with the aim to obtaining a contradiction, that $$\exists k \in \N$$ such that $$\psi: S \to \N_k$$ is a bijection. This would be possible if $$S$$ were finite.

Note that $$\forall j \in \N_k: j \in \N$$.

However, $$k \in \N$$ but $$k \notin \N_k$$ so $$\N_k \subset \N$$.

Consider the restriction $$\phi \restriction_{\N_k}$$ of $$\phi$$ to $$\N_k$$.

Suppose $$\phi \restriction_{\N_k}: \N_k \to S$$ is a bijection.

Consider $$\phi \left({k}\right) \in S$$.

Now $$\exists j \in \N_k: \phi \left({j}\right) = \phi \left({k}\right)$$.

But as $$j \ne k$$ it follows that $$\phi$$ can not be an injection.

So there can be no such $$k$$ and so $$S$$ is infinite.