# Product of perfectly-$T_3$ spaces

This is (sort of) a continuation of this question

Let $X$ be a topological space. Then, we say that $X$ is perfectly-$T_3$ iff it is $T_1$ and for every $x\in X$ and closed subset $C\subseteq X$ not containing $x$ there is a continuous function $f\colon X\rightarrow [0,1]$ such that $\{ x\} =f^{-1}(0)$ and $C=f^{-1}(1)$.

Is the product of perfectly-$T_3$ spaces necessarily perfectly-$T_3$?

Once again, I think the answer is "No.", but I was initially wrong last time, and that could very well be the case again . . .

Let $X$ be the double arrow space (i.e., the subspace of the lexicographically ordered square consisting of the top and bottom edges). As is shown at the link, $X$ is a perfectly normal compact Hausdorff space whose square is not hereditarily normal and hence is not metrizable. The main theorem here shows that $X$ does not have a $G_\delta$ diagonal. Since the diagonal is closed in $X^2$ but not a $G_\delta$, $X^2$ cannot be perfectly regular.