Let $A$ be an infinite set and let $B$ be a set such that $A$ is equinumerous to a subset of $B$. Then, $B$ is infinite.

Question

To me, the proof is as simple as this:

Let $C\subset B$ such that $A\sim C$. Then, as $A$ is infinite, we have that $C$ is infinite. Thus, as $C\subset B$, it must be that $B$ is infinite. Thus, $B$ is equinumerous to a subset of itself, and so $B$ must be infinite.

But that feels WAY too easy. Is it sufficient to say that because some set is equinumerous to an infinite set it must be infinite as well? If so, then the proof is quite quick.

Answer 1

"Is it sufficient to say that because some set is
equinumerous to an infinite set, it must be infinite as well?"

Yes, of course. Assume the opposite and use the
definition of "equinumerous". You easily get a contradiction.

And yes, the proof is easy since the statement is quite trivial.