I've already shown that $\operatorname{Cl}(A \times B)$ is a subset of $\operatorname{Cl}(A) \times \operatorname{Cl}(B)$, but I'm not sure how to show the other containment: $\operatorname{Cl}(A) \times \operatorname{Cl}(B)$ is a subset of $\operatorname{Cl}(A \times B)$.
I found a very neat proof that uses "closure points" but closure points are not something we have in our repertoire so I don't want to use it. I want to know if there's a simple way of proving this containment without using closure points.
Here's what I have so far:
I know that $\operatorname{Cl}(A) \times \operatorname{Cl}(B)$ is a closed set from the first containment. Since it is a closed set, it is contained within its closure... so I basically get that $\operatorname{Cl}(A) \times \operatorname{Cl}(B)$ is a subset of $\operatorname{Cl}(\operatorname{Cl}(A) \times \operatorname{Cl}(B)) = \operatorname{Cl}(A \times B)$. I think I'm on the right track but obviously this isn't exactly right.
Thank you for your insight.
Hint: Recall the definition of the product topology. Show that a point outside $\operatorname{Cl}(A \times B)$ is also outside $\operatorname{Cl}(A) \times \operatorname{Cl}(B)$.