A closed contained in an open set Let $X$ be a Hausdorff first countable topological space. Fix a countable decreasing local base of open neighborhood $(U_n:n\ge 1)$ of a point $x \in X$.

Question. Is it true that there exists always an integer $n$ and a closed set $F$ such that
  $$
U_n \subseteq F \subseteq U_1\,\,?
$$

 A: Too long for a comment.
Your question is equivalent to the existence of some $n$ such that $\overline{U_n}\subseteq U_1$,[ which in turn is equivalent (due to the decreasing nature of $U_n$) to $\partial U_n \subseteq U_1$.]
Now if $X$ is a topological group, then the claim is simple. Indeed, by the continuity of the addition, you can find an $n$ such that $U_n+U_n \subseteq U_1$. Then we have
$$\overline{U_n} \subseteq U_n+U_n \subseteq U_1$$
The same idea of proof works if the topopology is metrisable, and I suspect also if the topology is given by an uniformity.
A: As noted in comments, if $X$ is regular, then the answer is yes. But there are counterexamples in general.
The half-disk topology on $ℝ × [0, ∞)$ is a classical example of a Hausdorff non-regular space (https://topology.jdabbs.com/spaces/S000070). It is also first-countable. Now if we take just the subspace of rational points, i.e. $ℚ × ([0, ∞) ∩ ℚ)$, we obtain a countable space that also inherits all the properties mentioned above.
