We call $X$ strongly collectionwise normal if the set $\mathcal{U}_\Delta$ of all neighbourhoods of the diagonal $\Delta_X$ of $X\times X$ is a uniformity. This is equivalent to the property that for all $U\in\mathcal{U}_\Delta$ there exists $V\in\mathcal{U}_\Delta$ with $V\circ V\subseteq U$.

In the article Even covers and collectionwise normal spaces by Shapiro and Smith the following theorem is proven:

Theorem 4.8. If $X$ is a completely regular space then the following are equivalent:

1. $X$ is strongly collectionwise normal

2. Every even open cover of $X$ is normal

Here a cover $\mathcal{V}$ of $X$ is even if there is $W\in\mathcal{U}_\Delta$ such that $\{W(x) : x\in X\}$ refines $\mathcal{V}$, where $W(x) = \{y : (x, y)\in W\}$.

An open cover $\mathcal{V}$ is normal if there exists a sequence $(\mathcal{V}_n)$ of open covers such that $\mathcal{V}_{n+1}$ is a star-refinement of $\mathcal{V}_n$, and what authors of the article show to be equivalent to the property that $\mathcal{V}$ is $\aleph_0$-even, that is there exists a sequence $W_n\in\mathcal{U}_\Delta$ such that $W_{n+1}\circ W_{n+1}\subseteq W_n$ and $\{W_1(x) : x\in X\}$ is a refinement of $\mathcal{V}$.

The issue I have is with the proof that $(2)$ implies $(1)$. It is claimed that if $W\in\mathcal{U}_\Delta$ then $\mathcal{W} = \{W(x) : x\in X\}$ is an even cover, and that implies that there is $U\in\mathcal{U}_\Delta$ such that $U\circ U\subseteq W$. However, I don't see how it would imply this. Clearly there is a sequence $W_n\in\mathcal{U}_\Delta$ such that $W_{n+1}\circ W_{n+1}\subseteq W_n$ and $\{W_1(x) : x\in X\}$ refines $\mathcal{V}$. However, I don't think we can guarantee that there is $n$ with $W_n(x)\subseteq W(x)$ for all $x\in X$, which would imply that $W_n\subseteq W$, and hence allow us to take $U = W_{n+1}$.

So, can this proof somehow be fixed? And, what I am interested in the most, is if we can nonetheless show that a fully normal space is strongly collectionwise normal, even if this particular theorem is wrong.

We call $X$ strongly collectionwise normal if the set $\mathcal{U}_\Delta$ of all neighbourhoods of the diagonal $\Delta_X$ of $X\times X$ is a uniformity. This is equivalent to the property that for all $U\in\mathcal{U}_\Delta$ there exists $V\in\mathcal{U}_\Delta$ with $V\circ V\subseteq U$.

In the article Even covers and collectionwise normal spaces by Shapiro and Smith the following theorem is proven:

Theorem 4.8. If $X$ is a completely regular space then the following are equivalent:

1. $X$ is strongly collectionwise normal

2. Every even open cover of $X$ is normal

Here a cover $\mathcal{V}$ of $X$ is even if there is $W\in\mathcal{U}_\Delta$ such that $\{W(x) : x\in X\}$ refines $\mathcal{V}$, where $W(x) = \{y : (x, y)\in W\}$.

An open cover $\mathcal{V}$ is normal if there exists a sequence $(\mathcal{V}_n)$ of open covers such that $\mathcal{V}_{n+1}$ is a star-refinement of $\mathcal{V}_n$ and $\mathcal{V}_1 = \mathcal{V}$, and what authors of the article show to be equivalent to the property that $\mathcal{V}$ is $\aleph_0$-even, that is there exists a sequence $W_n\in\mathcal{U}_\Delta$ such that $W_{n+1}\circ W_{n+1}\subseteq W_n$ and $\{W_1(x) : x\in X\}$ is a refinement of $\mathcal{V}$.

The issue I have is with the proof that $(2)$ implies $(1)$. It is claimed that if $W\in\mathcal{U}_\Delta$ then $\mathcal{W} = \{W(x) : x\in X\}$ is an even cover, and that implies that there is $U\in\mathcal{U}_\Delta$ such that $U\circ U\subseteq W$. However, I don't see how it would imply this. Clearly there is a sequence $W_n\in\mathcal{U}_\Delta$ such that $W_{n+1}\circ W_{n+1}\subseteq W_n$ and $\{W_1(x) : x\in X\}$ refines $\mathcal{V}$. However, I don't think we can guarantee that there is $n$ with $W_n(x)\subseteq W(x)$ for all $x\in X$, which would imply that $W_n\subseteq W$, and hence allow us to take $U = W_{n+1}$.

So, can this proof somehow be fixed? And, what I am interested in the most, is if we can nonetheless show that a fully normal space is strongly collectionwise normal, even if this particular theorem is wrong.