Let be either the closed unit interval or . Let be any one of the following spaces:
The cross product is normal in all four cases. The factor, in the order listed, is compact, countably compact, paracompact and Lindelof. Thus they are all countably paracompact. According to the Dowker’s theorem, the product of a normal space and a compact metric space is normal if and only if is countably paracompact. My goal here is to show that the four cases of here cannot be hereditarily normal. This is from a theorem due to Katetov. We prove the following theorem.
Theorem. If is hereditarily normal, then either is perfectly normal or every countably infinite subspace of is closed and discrete.
Proof. Suppose that is not perfectly normal and has a countably infinite subset that is not closed and discrete. Let be a closed set that is not a set. Let be an infinite set with an accumulation point . We assume that .
We show that the open subspace is not normal. To this end, let and . The sets and are disjoint closed subspaces of the open subspace . Suppose we have disjoint open sets and such that and .
For each , let . Each is open and . Thus . Let . Then . This means . If , then (which is impossible). So we have , indicating that is a set, and leading to a contradiction. So the subspace is not normal.
Corollary. For countably compact spaces (in particular compact spaces) and , if the Cartesian product is hereditarily normal, then both and are prefectly normal.
Proof. Note that both factors, being countably compact, cannot have closed and discrete infinite subsets.
Comment. Note that the converse of the corollary is not true. Let both factors be the double arrow space, which is perfectly normal. But the square of the double arrow space contains a copy of the Sorgenfrey Plane, which is not normal.
[Katetov] Katetov, M.,  Complete normality of Cartesian products, Fund. Math., 36, 271-274.