Countable open covering of normal space

Question

I read the following claim in Z.Frolik's article "A generalization of realcompact spaces" on page 135.

Two subset $M$ and $N$ of a space $X$ are called completely seperated if there exists a real valued continuous function $f$ on $X$ with $f(M)\subset \{0\}$ and $f(N)\subset\{1\}$.

Claim: Let $X$ be a normal space. For every countable open covering $\mathfrak{U}$ of $X$, there exists a countable open covering $\mathfrak{B}$ of $X$ such that for every $B$ in $\mathfrak{B}$ there exists an $A$ in $\mathfrak{U}$ such that $B$ and $X-A$ are completely seperated.

I didn't show the proof of the claim.

Answer 1

The claim is false it would imply that normal spaces are countably paracompact and hence that normality of $X$ would imply normality of $X\times[0,1]$. The latter is not the case, see Mary Ellen Rudin, A normal space $X$ for which $X\times I$ is not normal, Fundamenta Mathematicae, 73 (1971/72), 179-186.

To show that the property in the claim implies countable paracompactness we use Theorem 5.2.1 in Engelking's General Topology. Let $\{U_n:n\in\omega\}$ be an increasing open cover; we need to find open $O_n$ such that $\operatorname{cl}O_n\subseteq U_n$ for all $n$ and $\bigcup_nO_n=X$. Take an open cover $\{V_m:m\in\omega\}$ as in the claim; hence for every $m$ an $n$ such that $\operatorname{cl}V_m\subseteq U_n$. Now define $O_n=\bigcup\{V_m:m\le n$ and $\operatorname{cl}V_m\subseteq U_n\}$; then $\operatorname{cl}O_n\subseteq U_n$ for all $n$ and the $O_n$ form a cover (if $x\in V_m$ and $\operatorname{cl}V_m\subseteq U_n$ then $x\in O_n$).