Solution: Let $x \in A$. Define a class function $F_x$ on $B$ by $F_x(y) = (x,y)$. The domain of $F_x$ is a set, so by replacement the range of $F_x$, which is equal to the Cartesian product $\{x\} \times B$, is also a set. Now define a class function $G$ on $A$ by $G(x) = \{x\} \times B$. (Note that this definition of $G$ only makes sense because we know that $\{x\} \times B$ is a set; a proper class cannot be the value of a function.) The domain of $G$ is a set, so by replacement the range of $G$ is also a set. By the union axiom $\bigcup \text{ran}(G)$ is a set and it remains to observe that $A \times B = \bigcup \text{ran}(G)$. Let $(x,y) \in A \times B$. Then $(x,y) \in \{x\} \times B \in \text{ran}(G)$, so $(x,y) \in \bigcup \text{ran}(G)$. Conversely, we have $\bigcup \text{ran}(G) \subset A \times B$ because every element of $\text{ran}(G)$ has the form $\{x\} \times B$ for some $x \in A$, and $\{x\} \times B \subset A \times B$.