Def: A space $X$ is regular if $\forall x \in X$, and for every closed set $C \subset F$ with $x \notin C$, there exists disjoint open sets $U$ and $V$ s.t $x \in U$ and $C \subset V$.
Problem: A subspace of a regular space is regular; $X \times Y$ is regular iff each of $X$ and $Y$ is regular.
So, the first part I think I figure out.
A Subspace of a regular space is regular.
Proof: Let $Y$ be a subspace of a regular space $X$ and let $x \in Y$ and $F$ be a closed set in $Y$ not containing $x$. Since $Cl_{X}(F) \cap Y = F$ then $x \notin Cl_{X}(F)$. Then because $X$ is regular $\exists U,V$ disjoint open set in $X$ such that $ x \in U$ and $Cl_{X}(F) \subset V$. Then $U \cap Y$ and $V \cap Y$ are open sets in Y such that $x \in U \cap Y$ and $F \subset V \cap Y$. Therefore $Y$ is a regular subspace of $X$.
The second half of the question is where I'm having trouble. I feel like it should follow straight forward from the definition of a regular space and work out component wise on $X \times Y$, but at the same time I know how these questions go and that the question wouldn't be phrased how it is if I didn't need to use the fact that a subspace of a regular space is regular. Also, talking to a friend made me feel this to be even more true, but I can't seem to get it or see why it is really needed. Here's my attempt though.
$X \times Y$ is regular iff each of $X$ and $Y$ is regular.
Proof:
$\Rightarrow$) Assume $X \times Y$ is regular. Then $\forall (x,y) \in X \times Y$ and for every closed $(F_{1}, F_{2}) \subset X \times Y$ with $(x,y) \notin (F_{1}, F_{2})$ there exists disjoint open sets $(U_{1}, U_{2}), (V_{1}, V_{2}) $in $ X \times Y$ such that $(x,y) \in (U_{1}, U_{2})$ and $(V_{1}, V_{2}) \subset (F_{1}, F_{2})$. Then $x \in U_{1} \subset X$, $y \in U_{2} \subset Y$, and $V_{1} \subset F_{1}$ in $X$ and $V_{2} \subset F_{2}$ in $Y$. Therefore, $X$ is regular and $Y$ is regular.
$\Leftarrow$) I'm not really sure here, and the proof I have now for this part is fairly analogous to the first implication.
As far as using the subspace result for proving the second half of the question the only thing I can really think to do is to consider, the subspaces $X$ and $Y$ in $X \times Y$ but I wasn't able to quite get the proof when messing around with that and my notation is probably a little off since I have not done much with products in topology, so I'll spare everyone and not post that. This question seemed like it would be fairly easy on first glance I'm not sure why it's getting me so confused. Anyways, thank you very much for any help/insight you can provide.
Best Answer
Here are some suggestions; I’ve left a lot of details for you to complete.
The product question is much easier if you first prove this little
For the proof of $(\Leftarrow)$, suppose that $\langle x,y\rangle\in X\times Y$ and $U$ is an open set in $X\times Y$ such that $\langle x,y\rangle\in U$. Then by the definition of the product topology there are open sets $V\subseteq X$ and $W\subseteq Y$ such that $\langle x,y\rangle\in V\times W\subseteq U$. Clearly $x\in V$ and $y\in W$, so you can use the regularity of $X$ and $Y$ and the proposition to get open sets $G\subseteq X$ and $H\subseteq Y$ such that $x\in G\subseteq\operatorname{cl}_X G\subseteq V$ and $y\in H\subseteq\operatorname{cl}_Y H\subseteq W$. Now consider the open set $G\times H$, which clearly contains $\langle x,y\rangle$; what is its closure in $X\times Y$?
For the proof of $(\Rightarrow)$ you need to assume that $X\times Y$ is regular and then prove that $X$ and $Y$ are regular. To show directly that $X$ is regular, you must start start with a point $x\in X$, not a point in $X\times Y$. If you’re going for a direct proof, and you use the proposition, you’ll also start with an open set $U$ containing $x$. Let $y$ be any point of $Y$, and consider the open neighborhood $U\times Y$ of $\langle x,y\rangle$. Get an open set $V\subset X\times Y$ such that $$\langle x,y\rangle\in V\subseteq\operatorname{cl}_{X\times Y}V\subseteq U\times Y\;,$$ and find a basic open set in the product topology (i.e., an open ‘box’, like $V\times W$ in the proof of $(\Leftarrow)$) containing $\langle x,y\rangle$ and contained in $V$. You should have little trouble using that ‘box’ to get an open set in $X$ containing $x$ whose closure is contained in $U$.
However, for this direction you can, as you suspected, use the first part of the problem. Pick any $y\in Y$; then $X\times \{y\}$ is a subspace of the regular space $X\times Y$, so it’s regular, and it’s also homeomorphic to $X$, so $X$ is regular as well. You can handle $Y$ similarly.