Existence of finite sets of infinite set without using AC

Question

Is it possible to prove that every infinite set $B$ has a subset of cardinality $n$, for every natural $n$, without using AC? I know how to prove this claim by induction. In the induction step I chose arbitrarily one element of $B$. My question is, does this count as using AC? if not, why?

Answer 1

No, this is not using the axiom of choice.

This is just instantiating an existential quantifier. Since the induction hypothesis says that there is some $B_n\subseteq A$ and $B_n$ has $n$ elements, but since $A$ is infinite, $A\nsubseteq B_n$, therefore $\exists a(a\in A\setminus B_n)$.

Think about the proof by induction as proving a schema of proofs, which tells you for every finite number something holds. This is not quite true (but not quite false either), but it helps putting the finger on why the axiom of choice is not needed here.