On the small first countable regular spaces Here are some definitions.

A  countably  infinite closed discrete set $A \subset  X$
  has  property  $D$  in $X$  provided  there exists a  discrete family 
  of open sets $\{U_a :  a \in A\}$ such that $U_a \cap A = \{a\} $ for all $a\in  A$.
A  space  $X $ is said to have  property  $D$  provided  that every countably  infinite closed discrete set in $X$  has  property $D$    in $X$. 
A space is called pseudo-normal if every countable closed subset has arbitrarily small closed nbhds. 

It is known each pseudonormal space has property $D$.
It is said in the Handbook of set-theoretic topology page 155: The following result tells that the small first countable regular spaces are pseudo-normal;


Question: how to see that the small first countable regular spaces are pseudo-normal from Theorem 12.2?

Thanks for your help.
 A: $\newcommand{\cl}{\operatorname{cl}}$When Eric says that ‘small first countable regular spaces are pseudonormal’, he’s talking about spaces $X$ such that $|X'|<\mathfrak b$, not just spaces that are themselves of cardinality less than $\mathfrak b$. Specifically, he’s referring to the fact that $\mathfrak{b}_p=\mathfrak b$. The reason that $\mathfrak{b}_p$ is defined in terms of $|X'|$ instead of $|X|$ is that if $X$ is pseudonormal, you can add any number of isolated points to $X$, and the new space will still be pseudonormal. And since regularity was mentioned in the comments, note that the proof of Theorem $12.2$ uses the regularity of $X$.
By the way, there are several confusing typos in the proof that $\mathfrak b\le\mathfrak{b}_p$. Here, with slightly different notation in places, is a corrected version.

We prove that if $X$ is a first countable regular space with $|X'|<\mathfrak b$, then $X$ is pseudonormal. Let $F$ be a countable closed subset of $X$, and let $U$ be a nbhd of $F$; we must find a nbhd $V$ of $F$ such that $\cl V\subseteq U$. This is trivial if $F$ is finite — note the use of regularity here — so assume that $F=\{x_k:k\in\omega\}$. For each $k\in\omega$ let $\mathscr{B}_k=\{B_k(n):n\in\omega\}$ be a nbhd base at $x_k$ such that $B_k(n)\supseteq B_k(n+1)$ for each $n\in\omega$, and $\cl B_k(0)\subseteq U$.
For $f\in{}^\omega\omega$ let $B(f)=\bigcup_{k\in\omega}B_k\big(f(k)\big)$; then $\{B(f):f\in{}^\omega\omega\}$ is a nbhd base for $F$ in $X$. For each $y\in X'\setminus U$ there is an $f_y\in{}^\omega\omega$ such that $y\notin\cl B(f_y)$. Now $|X'\setminus U|\le|X'|<\mathfrak b$, so $\{f_y:y\in X'\setminus U\}$ is bounded in $\langle{}^\omega\omega,\le^*\rangle$, and there is therefore a $g\in{}^\omega\omega$ such that $f_y\le^* g$ for all $y\in X'\setminus U$. $B(g)$ is a nbhd of $F$, and I claim that $\cl B(g)\subseteq U$.
To see this, suppose that $y\in X'\setminus U$. Then there is an $m\in\omega$ such that $f_y(k)\le g(k)$ for all $k\ge m$, and $$\cl B(g)\subseteq\cl\left(B(f_y)\cup\bigcup_{k<m}B_k(0)\right)=\cl B(f_y)\cup\bigcup_{k<m}\cl B_k(0)\;.$$ By construction $y\notin\cl B(f_y)$ and $\bigcup_{k<m}\cl B_k(0)\subseteq U\subseteq X\setminus\{y\}$, so $y\notin\cl B(g)$. Thus, $\cl B(g)\cap(X'\setminus U)=\varnothing$. Finally, $B(g)\subseteq U$, and every point of $(X\setminus X')\setminus U$ is isolated, so $\cl B(g)\cap(X\setminus U)=\varnothing$, and $\cl B(g)\subseteq U$, as desired. $\dashv$.

A: If $X$ is a first countable regular space so small that $|X|<{\frak b}_p$, then $X$ is pseudo-normal by the second row of Theorem 12.2. Is this OK?
