Normal countably compact implies collectionwise normal without T1? In this page of Dan Ma's topology blog about collectionwise normal spaces he proves this result:

Proposition: Any (Hausdorff) normal and countably compact space is collectionwise normal.

The blog assumes that spaces are Hausdorff (or T1 here), but I am interested to know what happens without the T1 assumption.
More specifically, the proposition follows from the following:

Lemma: If $X$ is a T1 space, the following are equivalent:

*

*(A) $X$ has countable extent;

*(B) All discrete families of nonempty closed subsets of X are at most countable.


Here the extent of a space $X$ is the supremum of the cardinalities of closed discrete subsets of $X$.  A discrete family of subsets of $X$ is a family such that each point of $X$ has a nbhd meeting at most one set in the family.
The proof of the lemma is not difficult (see below for completeness).  In fact, (B) implies (A) always, even without the T1 assumption.

Question: Does (A) imply (B) without the T1 assumption?

I can't prove it, but I don't see a counterexample either.  If it were true, we could generally conclude that every normal limit point compact space is collectionwise normal (since limit compact spaces have countable extent).  For T1 spaces it is not a generalization since limit compact is equivalent to countably compact in that case.

Proof of (A) implies (B) assuming T1:  Let $\mathscr{F}$ be a discrete family of nonempty closed subsets.  For each $F\in\mathscr{F}$ pick some $x_F\in F$.  Then $\mathscr{G}=\{\{x_F\}:F\in\mathscr{F}\}$ is a discrete family of closed subsets (singletons).  Hence $A=\cup\mathscr{G}=\{x_F:F\in\mathscr{F}\}$ is closed and discrete.  Hence $A$ is at most countable and the same is true of $\mathscr{F}$.
Proof of (B) implies (A) without extra assumption: Let $A$ be a closed and discrete subset of $X$.  For any $x\in A$, the singleton $\{x\}$ is closed in $A$ because $A$ is discrete, and $A$ is closed in $X$, therefore $\{x\}$ is closed in $X$.  So the family $\mathscr{F}=\{\{x\}:x\in A\}$ is a discrete family of nonempty closed subsets of $X$.  Hence $\mathscr{F}$ is at most countable and so is $A$.
 A: Brian has already shown that the lemma (A) implies (B) is false if one does not assume $T_1$.
Here is a proof that the result in the original title is true, bypassing the lemma entirely.

Proposition: (without assuming $T_1$) Any normal countably compact space $X$ is collectionwise normal.

Proof: Take a discrete family of closed subsets of $X$.  Since $X$ is countably compact, the family must be finite.  This is Lemma 2 in this answer.  Now we have a finite number of pairwise disjoint nonempty closed subsets, and we can enclose each set in the family in a open set with the open sets pairwise disjoint, by normality of $X$.
A: For $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k<n\}$, and let $Y$ be $\Bbb N$ with the topology
$$\{U_n:n\in\Bbb N\}\cup\{\Bbb N\}\,,$$
and let $D$ be an uncountable space with the discrete topology. Let $X=D\times Y$; clearly
$$\mathscr{F}=\big\{\{x\}\times Y:x\in D\big\}$$
is an uncountable discrete family of closed sets in $X$. Let $A\subseteq X$. If $|A\cap(\{x\}\times Y)|>1$ for some $x\in D$, then $A$ is not discrete, and if $|A\cap(\{x\}\times Y)|=1$ for some $x\in D$, then $A$ is not closed in $X$, so $X$ has no non-empty closed discrete subsets.
