Does every neighborhood of a closed set $F$ in a metric space contain an $\varepsilon$-neighborhood of $F$? Let $(X, d)$ be a metric space, $F$ a closed subset of $X$, and $U$ an open neighborhood of $F$. Define the distance from a point $x \in X$ to $F$ as
$$
d(x, F) = \inf_{y \in F} d(x, y),
$$
and define the $\varepsilon$-neighborhood of $F$ to be
$$
B_d(F, \varepsilon) = \{ x \in X : d(x, F) < \varepsilon\}.
$$
Is it true that the $\varepsilon$-neighborhood of $F$ is contained in $U$ for some sufficiently small $\varepsilon$?

Not in general, as the suggested counterexample shows, but it works if $F$ is compact, because $g(x)=d(x,U^c):F\to \mathbb{R}$ is continuous (actually 1-Lipschitz by triangle inequality) and strictly positive, therefore it has a positive minimum.
