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
Proposition: A space $X$ is regular if and only if for each point $x\in X$ and each open set $U$ containing $x$ there is an open set $V$ such that $x\in V\subseteq\operatorname{cl}_X V\subseteq U$.
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.
Yes, your proof is correct. However, we can even say more: The product $X=\prod_i X_i$ of a arbitrary family of completely regular spaces $(X_i)_i$ is completely regular. This can be proven by a generalization of your approach. Another approach is via initial topologies. Note that a space is completely regular if and only if it has the initial topology with respect to some family of real-valued continuous maps. Since the product $X$ has the initial topology with respect to all projections $p_i:X\to X_i$, it has the initial topology with respect to a family of continuous real-valued maps $X\to X_i \to \Bbb R$ (see for example the Transitive law of initial topologies near the end of Henno Brandsma's answer here), and is thus completely regular.
Best Answer
There are multiple mistakes in your proof. First of all, the set $U$ is a subset of $X$, not a a subset of $X_i$. This mistake also cannot be fixed by considering the projection of $\pi_i(U)$ onto $X_i$, as $\pi_i(x)$ may be inside of $\pi_i(U)$. For example, consider a circle in $\mathbb{R}^2$ and the origin.
Also, I assume you want to consider the product topology on the product of spaces. However, the infinite product of open sets is in general not open in the product topology (see also Box Topology).
If you want a hint as to how to prove the assertion, take a look at this answer on MO.