I'm having trouble proving that if $A$ is an infinite set, then $A\approx A\times A$ using Zorn's Lemma directly.
My problem is I don't know how to choose the correct set to which apply Zorn; I have tried $$X=\{f_{B}\colon B\times B\longrightarrow B\mid B\subseteq A,\ f_{B}\ {\rm injective}\}$$ since it's obvious that $A\preceq A\times A$, and with $(X,\leq)$ (the order $f_{B}\leq f_{B'}$ if $B\subseteq B'$ and $f_{B'}$ restricted to $B\times B$ is $f_{B}$) I manage to get a maximal function of that kind, but then I can't "expand it" as usual to prove it must have domain $A\times A$. Do you know which set $X$ should I work with?
So, you're ordering such injective functions $f_B: B \times B \rightarrow B$ by the relation that $f_B \leq f_{B'}$ if and only if $B \subseteq B'$ and $f_{B'}$ restricts to $f_B$ on $B \times B$, right?
By Zorn's lemma, there is a maximal such injective function $f_B: B \times B \rightarrow B$. If $B$ is properly contained in $A$, there are two cases (i): $A -B$ is finite or (ii): $A-B$ is infinite. In the first case, you're already done, because $B$ and $A$ would have the same cardinality. In the second case, you have plenty of leeway to extend $f_B$ to an injective function on $C \times C \rightarrow C$, where $C - B$ is infinite.