By preserved by products I mean - $\prod X_{\alpha}$ has property $P$ iff $X_{\alpha}$ has property $P$ for all $\alpha$ in index set
Also, $X$ is Completely Hausdorff if for $x\neq y$ in $X$, $\exists$ continuous function $f:X\to I$ with $f(x) = 0$, $f(y) = 1$.
Are Semiregularity and Completely Hausdorff preserved by products? If not, then is any direction of the iff statement above true?
Suppose that $X_\alpha$ is completely Hausdorff for each $\alpha\in A$, and let $x,y\in X$ be distinct points; there is an $\alpha\in A$ such that $x_{\alpha}\ne y_{\alpha}$, and there is continuous $f_\alpha:X_\alpha\to[0,1]$ such that $f_\alpha(x_\alpha)=0$ and $f_\alpha(y_\alpha)=1$. Now define
$$f:X\to[0,1]:z\mapsto f_\alpha(z_\alpha)\;;$$
if $\pi_\alpha:X\to X_\alpha$ is the projection map, $f=f_\alpha\circ\pi_\alpha$. Clearly $f$ is continuous, $f(x)=0$, and $f(y)=1$. Thus, $X$ is completely Hausdorff.
Conversely, if $X$ is completely Hausdorff and non-empty, then each $X_\alpha$ is completely Hausdorff: complete Hausdorffness is evidently hereditary, and if we fix $x\in X$, the subset
$$\big\{y\in X:y_\beta=x_\beta\text{ for all }\beta\in A\setminus\{\alpha\}\big\}$$
of $X$ is homeomorphic to $X_\alpha$.
Now suppose that each $X_\alpha$ is semiregular, and let $\mathscr{B}_\alpha$ be a base of regular open sets for $X_\alpha$. Then $X$ has a base $\mathscr{B}$ whose elements are the sets, $\prod_{\alpha\in A}U_\alpha$ such that $U_\alpha=X_\alpha$ for all but finitely many $\alpha\in A$, and $U_\alpha\in\mathscr{B}_\alpha$ whenever $U_\alpha\ne X_\alpha$. Let $B=\prod_{\alpha\in A}U_\alpha\in\mathscr{B}$, and let $F=\{\alpha\in A:U_\alpha\ne X_\alpha\}$. It’s easy to check that the sets $\pi_\alpha^{-1}[U_\alpha]$ for $\alpha\in F$ are regular open in $X$; $B=\bigcap_{\alpha\in F}\pi_\alpha^{-1}[U_\alpha]$, and the intersection of finitely many regular open sets is regular open, so $B$ is regular open, and $X$ is semiregular.
I am not at the moment sure about the other direction, since semiregularity is not hereditary.