A Cartesian product of metric spaces is perfectly $\kappa$-normal

A space $X$ is called perfectly $\kappa$-normal if the closure of any open set (that is, every canonical closed set) is a zero-set.

How can i prove this proposition directly?

$Proposition$: A Cartesian product of metric spaces is perfectly $\kappa$-normal.

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Do you mean "is the zero set of some continuous function"? – Rhys Nov 21 '12 at 8:37
@Rhys: Yes. Such spaces have also been called Oz-spaces (or $O_z$-spaces). – Brian M. Scott Nov 21 '12 at 8:40
I’ve never seen Ščepin’s proof of the result, and I don’t see a proof at the moment. However, I did find this paper, which contains a proof of a more general result; see Theorem $2$. – Brian M. Scott Nov 21 '12 at 9:53
This is equivalent to proving that every metric space is perfectly $\kappa$-normal, right? Having seen Brian's comment, I guess it isn't this easy, but my approach would have been, given some open set $U \subset (X,d_X)$, to define $f_U : X \to \mathbb{R}^+ ~,~ f(x) := \text{inf}\{d_X(x,y) ~|~ y \in U\}$. Then show that $f$ is continuous... – Rhys Nov 21 '12 at 10:19
@Rhys: No, it isn’t: products of uncountably many non-trivial metric spaces aren’t metrizable, and the theorem is for arbitrary products. It is fairly trivial that metric spaces are $\kappa$-normal. – Brian M. Scott Nov 21 '12 at 10:50