Is the Axiom of Choice needed to prove the following: If $|A|>|B|$, then there exists a $C\subseteq A$ such that $|C|=|B|$?
I'm thinking that AC isn't needed. If $|A|>|B|$, then there exists an injective function $f$ from $B$ into $A$ and there doesn't exist an injective function from $A$ into $B$. Let $C\subseteq A$ be the image of $f$. Then, $f$ is both an injective and surjective function from $B$ to $C$. This means that $|C|=|B|$. Is this right?
In fact, we don't need the strictness of the inequality at all.
As you observe, by definition $\vert A\vert\ge \vert B\vert$ iff there is some injection $f:B\rightarrow A$. The corestriction of $f$ to its image $$\hat{f}:B\rightarrow im(f):b\mapsto f(b)$$ is trivially a bijection, and its image $im(f)=im(\hat{f})$ is trivially a subset of $A$, so we can just take it to be our $C$.
(Note that I'm treating $f$ and $\hat{f}$ as different objects. This reflects the majority approach to functions outside of set theory, namely that a function does come equipped with a specific codomain. By contrast, inside set theory the majority approach is to view a function as simply a set of ordered pairs, so "$f$ is a bijection from $B$ to $im(f)$" is literally true.)