Rudin's Dowker space is collectionwise normal I am trying to follow K.P. Hart's course Set-theoretic methods in general topology. In Chapter 6, Rudin's Dowker space $X$ is defined as follows. Let $P=\prod_{n=1}^\infty(\omega_n+1)$ be the box product of the successors of the first $\omega$-many uncountable ordinals, let $X'=\{x\in P:(\forall n)\,\operatorname{cf} x_n>\omega\}$, and let $X=\{x\in X':(\exists i)(\forall n)\ \operatorname{cf}x_n<\omega_i\}$. Exercise 2.8 asks to show that $X$ is collectionwise normal, using the following hint: prove that if $\mathcal{F}$ is a discrete family of closed subsets of $X$ then $\mathcal{F}'=\{\operatorname{cl}_{X'}F:F\in\mathcal{F}\}$ is a discrete family of closed subsets of $X'$. 
It was already proved that disjoint closed sets in $X$ have disjoint closures in $X'$. I can prove that the space $X'$ is collectionwise normal, and using the statement of the hint I can also show that $X$ is collectionwise normal. But somehow I am not able to prove the hint.
Question: Is it true that if $Y\subseteq Y'$ are topological spaces, disjoint closed subsets of $Y$ have disjoint closures in $Y'$, and $\mathcal{F}$ is a discrete family of closed subsets of $Y$, then $\mathcal{F}'=\{\operatorname{cl}_{Y'}F:F\in\mathcal{F}\}$ is a discrete family of closed subsets of $Y'$? We can also assume that $Y$ and $Y'$ are normal. Or is that a special property of the above spaces $X$, $X'$?
 A: The hint is a classical fact. No need for any assumptions on $X$. So suppose $\mathcal{F}$ is a discrete family. To see that $\{\overline{F}: F \in \mathcal{F}\}$ is also discrete, let $x \in X$ . Then $x$ has an open neighbourhood $O_x$ such that $\left|\{ F \in \mathcal{F}: O_x \cap F \neq \emptyset\}\right| \le 1$. But $O_x \cap F = \emptyset$ implies $O_x \cap \overline{F} = \emptyset$ (as for any $p \in O_x$, $O_x$ is a witnessing neighbourhood that $p \notin \overline{F}$), and this means that $O_x \cap F \neq \emptyset$ iff $O_x \cap \overline{F} \neq \emptyset$, so $O_x$ also intersects at most one set of the form $\overline{F}$, $F \in \mathcal{F}$, i.e.the family of closures is discrete. 
A: I think that I finally got it. No assumptions on the spaces are needed.
Spaces $X$, $Y$ are not related to the spaces mentioned in the question.
Lemma 1.
Let $\mathcal{F}$ be a discrete family of closed subsets
of a topological space $X$.
Then $\bigcup\mathcal{F}$ is closed.
Proof.
If $x$ belongs to the closure of $\bigcup\mathcal{F}$
then every neighborhood of $x$ intersects at least one
set $F\in\mathcal{F}$.
Since $\mathcal{F}$ is a discrete family,
there exists an open neighbourhood $U$ of $x$
and a set $F\in\mathcal{F}$ such that $U\cap G=\emptyset$
for every $G\in\mathcal{F}\setminus\{F\}$.
It follows that $U\cap V\cap F\neq\emptyset$ for any open
neighbourhood $V$ of $x$, hence $x$ belongs to the closure
of $F$.
Since $F$ is closed, we have
$x\in F\subseteq \bigcup\mathcal{F}$.
Lemma 2.
Let $X$ be a subspace of a topological space $Y$
such that any two closed disjoint subsets of $X$ have
disjoint closures in $Y$.
Let $\mathcal{F}$ be a discrete family of closed subsets
of $X$.
Then $\mathcal{F}'=\{\operatorname{cl}_Y(F)\!:F\in\mathcal{F}\}$ is a discrete family of closed subsets of $Y$.
Proof.
Let $y\in Y$.
We have to find an open neighbourhood of $y$ that
intersects at most one member of $\mathcal{F}'$.
If $y\in\bigcup\mathcal{F}'$ then there exists
$F\in\mathcal{F}$ such that $y\in\operatorname{cl}_Y(F)$.
Let us denote $\mathcal{G}=\mathcal{F}\setminus\{F\}$.
Then $\mathcal{G}$ is a discrete family of closed subsets
of $X$.
By the above lemma, $\bigcup\mathcal{G}$ is closed in $X$.
Since $F$ and $\bigcup\mathcal{G}$ are disjoint closed
subsets of $X$, their closures in $Y$ are disjoint,
hence $y\notin\operatorname{cl}_Y(\bigcup\mathcal{G})$.
The set
$U=Y\setminus\operatorname{cl}_Y(\bigcup\mathcal{G})$
is an open neighbourhood of $y$ that is disjoint to
every set $\operatorname{cl}_Y(G)$ for $G\in\mathcal{G}$,
hence $U$ intersects at most one member of $\mathcal{F}'$.
If $y\notin\bigcup\mathcal{F}'$ then
$Y\setminus\bigcup\mathcal{F}'$ is an open neighborhood
of $y$ that is disjoint to any member of $\mathcal{F}'$.
