Let $A$ and $B$ be infinite sets such that there is a surjective function $$ g: A \to B $$ Use Axiom of Choice to show that there is a set $ A' \subset A $ such that $ card (B) = card(A') \leq card(A) $
In my proof, I wanted to use axiom of choice to have $ f(A') = \max (A') $ This would mean I could have $ card(A') \leq card(B) $ and $ card(A') \geq card(B) $. This seems really wrong as $A'$ will have to be infinite and therefore doesn't have a maximum element by which I can constitute $f(A')$.
If $B$ is not ordered, there is no maximum element to talk about. Even if it is ordered, as you said, there might not be a maximum element (e.g. $\mathbb{Z}$).
For each $b \in B$, since $f$ is surjective $f^{-1}(b)$ is non-empty. This means that $\prod_{b \in B}f^{-1}(b)$ is a Cartesian product of non-empty sets, so by AC is itself non-empty. Let $g \in \prod_{b \in B} f^{-1}(B)$ (so $g$ is a function mapping each $b$ to an element in $f^{-1}(b)$), and we have $g$ is an injective (why?) function from $B$ to $A$. Set $A'$ to be the image of $g$ and the result follows.