Constructing sequence of nested closed sets in separable $T_{3\frac 1 2}$ space

Question

Let $X$ be a separable $T_{3\frac 1 2}$ space. In other words, for every closed $A \subset X$ and every $x \in X \setminus A$, there is a continuous function $f : X \to [0,1]$ for which $f(x) = 0$ and $f|_A =1$. In particular, $X$ is regular.

My question: Given a closed $A \subset X$, I want to construct a sequence of open sets $\{B_n\}_{n \geq 1}$, for which $B_1 \supset B_2 \supset B_3 \supset \cdots$, and for which $A = \mathop{\bigcap}_{n \geq 1} B_n$. Is this possible?

What I've tried:

Since $X$ is separable, let $Q = \{q_n\}_{n \geq 1} \subset X \setminus A$ be a countable dense subset; since $X$ is regular, for each $n$, there are open sets $U_n, V_n \subset X$ for which $q_n \in V_n$, $A \subset U_n$, and $U_n \cap V_n = \emptyset$. The first thing I thought to do was take $B_n = \mathop{\bigcap}_{i=1}^n U_i$. Clearly $A \subset \mathop{\bigcap}_{n \geq 1}B_n$, but we don't have equality; it could be that all of the open sets $V_n \ni q_n$ somehow "miss" some points outside of $A$, for example.

So maybe we can find a way to ensure $\{V_n\}_{n \geq 1}$ covers $X \setminus A$. Given $x \in X \setminus A$, denoting $V_x, U_x \subset X$ to be disjoint open sets with $x \in V_x$ and $A \subset U_x$, we clearly have $\{V_x\}_{x \in X \setminus A}$ is an open cover of $X \setminus A$, but separable and $T_{3\frac 1 2}$ does not imply Lindelöf, so we can't extract a countable subcover of $X \setminus A$.

Any suggestions for how to proceed? Or is this statement not true to begin with?

Answer 1

That's trivial: $B_n := A$ for each $n$. Or did you perhaps want the $B_n$'s to be open rather than closed? Then this is equivalent for the space to be perfect (i.e., closed sets are $G_\delta$). Then it is false, since there are lots of separable Tychonoff spaces, which are not perfect, for instance the Stone-Cech compactification of the integers.