I've tried to prove this using proof by contradiction:
Claim: $|A \times B| \lt |B|$.
Let $B$ be a subset of an infinite set $A$ and $f: B \to B \times B$ be a surjective function. Then $|B| \le |B \times B| \le |B| \le |A|$. In this case, since the cardinality of B is less or equal to A, the product of the two will be more than B itself. Therefore, our claim is false and the original claim is true.
Is this a proper approach to this question? Is there a better, more direct proof that I can use?