If $\prod X_i$ is regular, then each $X_i$ is regular.
Let $k \in \Bbb{N}$. Assuming no factor is empty, there exists at least one $x_i \in X_i$. Let $C$ be closed in $X_k$ and $x \notin C$. Then $p =(p_i)$ defined as $p_i = x_i$ if $i \neq k$ and $p_i = x$ if $i=k$ is not in the closed $\pi_k^{-1}(C)$. Since the product is regular, there exist disjoint open sets $U$ and $V$ separating $p$ and $\pi^{-1}_k(C)$. Moreover, there exist a basis element $\prod_i U_i$ containing $p$ and contained in $U$.
Now, I want to say that $\pi_k^{-1}(C) \subseteq V$ implies $V = \prod V_i$, where $V_i = X_i$ if $i \neq k$ and $V_k$ is some (proper) open set containing $C$, but I'm not sure how to show this. If I could show this, then $\emptyset = \prod U_i \cap \prod V_i = \prod (U_i \cap V_i)$ would have to imply $U_k \cap V_k = \emptyset$, which would finish the proof. Otherwise I am not sure how to finish the problem and could use a hint.
IMHO the easiest way to see your implication is to see that each $X_i$ is a subspace of the product and regularity is hereditary:
Pick a fixed $q = (q_i) \in \prod_{i \in I} X_i$ and then define for some $k \in I$:
$e_k : X_k \to \prod_i X_i$ by $e(x)_k = x$ ,$e(x)_i = q_i$ for $i \neq k$. Then $e_k$ is 1-1 continuous (as the compositions $\pi_i \circ e_k$ are either constant maps with value $q_i$ or the identity on $X_k$, so continuous always. And functions into the product that have continuous compositions with all projections are continuous (by a standard characterisation of the product topology)).
Also, $\pi_k$ is the continuous inverse, so $e_k$ is a homeomorphism between $X_k$ and $e_k[X_k] \subseteq \prod_i X_i$. And as said, subspaces of regular spaces are regular, hence so is $e_k[X_k]$ and thus $X_k$.