Suppose $X$ is any first countable regular space, and $D$ is the countable closed discrete subset of $X$, then could $D$ have arbitary small closed nbhd, i.e., for any open set $U$ of $X$ which contains $D$, $D$ has an small open set $V$, which satisfies that $\overline{V}\subset U$?
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Here’s a counterexample. Let $\mathscr{A}$ be a maximal family of almost disjoint subsets of $\omega$; i.e., if $A,B\in\mathscr{A}$ and $A\ne B$, then $A\cap B$ is finite. Let $X=\mathscr{A}\cup\omega$. Points of $\omega$ are isolated. Basic open nbhds of $A\in\mathscr{A}$ are sets of the form $$B_n(A)=\{A\}\cup\{k\in A:k\ge n\}\;.$$ (This is a Mrówka $\Psi$-space.) Clearly $X$ is a first countable, zero-dimensional Tikhonov space, and $\mathscr{A}$ is a closed, discrete set in $X$. Fix a countably infinite $D=\{A_n:n\in\omega\}\subseteq\mathscr{A}$, and let $U=D\cup\omega$. Let $V$ be any open nbhd of $D$ such that $V\subseteq U$. There is a function $n:\omega\to\omega$ such that $B_{n(k)}(A_k)\subseteq U$ for each $k\in\omega$ and the sets $B_{n(k)}(A_k)$ are pairwise disjoint. Let $S\subseteq\omega$ be such that $|S\cap B_{n(k)}(A_k)|=1$ for each $k\in\omega$; by the maximality of the family $\mathscr{A}$ there is an $A\in\mathscr{A}$ such that $A\cap S$ is infinite, and it follows immediately that $A\in\operatorname{cl}V$. Thus, $D$ has no open nbhd whose closure is contained in $U$. |
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