If it was given that both sets $A$ and $B$ are nonempty, then the answer would be yes. Indeed, if we assume that $A\times B$ is finite, then there is an injective function $f$ from $A\times B$ to $S_{n}$ for some $n\in\mathbb{Z}_{+}$ (Corollary 6.7). And if $B$ is nonempty, then for a fixed $b\in B$ , the restriction of $f$ on $A\times\{b\}$ is an injective function from $A\times\{b\}$ to $S_{n}$ , but then there is an obvious bijective correspondence of $A$ and $A\times\{b\}$ , the composite of which with $f$ proves $A$ to be finite (Corollary 6.7). Similarly, if $A$ is nonempty, then $B$ is finite. But if sets can be empty, then, as it was noted by Fran in the comments below, their product may be empty (finite) while one set is empty and the other is infinite: $\mathbb{R}\times\emptyset=\emptyset$ .