Condition on $F$ such that $U^\prime \subseteq U\cap F$ Let $X$ be a metric space. What condition should be assumed on a closed set $F$ such that, for every non-empty open set $U$, there exists another non-empty open set $U^\prime$ such that
$$
U^\prime \subseteq U\cap F?
$$
(I mean non-trivial condition: e.g. avoid "set $X=F$".)
 A: This cannot be done, for suppose $F$ is non-empty closed and proper, set $U = X\setminus F$ and then $U \cap F =\emptyset$ contains no non-empty open (open) set...
So only $F=X$ is possible.
A: Let $F$ be a non-trivial closed set. Since for every non-empty  open set $U$, there exists a non-empty open set $U^\prime$ such that $U^\prime\subseteq U\cap F$ we must have $U\cap F\neq \emptyset$. That is for every non-empty open set $U$, $U\cap F\neq \emptyset$ which is impossible since $X$ is normal (any metric space is normal). So we can't have what you want.
A: As stated, we need $F=X$ (see Henno Brandsma's answer). However, what if we only consider open sets $U$ that intersect $F$?
From $U' \subseteq U\cap F$ we know that $U' \subseteq F^\circ$, where $F^\circ:=\bigcup\{A\subseteq F \mid A\ \text{is open}\}$ denotes the interior of $F$. This means that every open set intersecting $F$ must also intersect the interior of $F$. A class of closed sets satisfying this condition are called regular closed sets.
A set $A$ is a regular closed set if $A = \overline{A^\circ}$. We prove that a closed set satisfies your (modified) condition if and only if the set is regularly closed.
Proof:
If $F$ is regular closed and $U\cap F\ne\emptyset$, then $U\cap\overline{F^\circ}\ne\emptyset$ and consequently $U\cap F^\circ \ne\emptyset$. Taking $U':=U\cap F^\circ$ gives a nonempty open set satisfying $U'\subseteq U\cap F$.
Conversely, suppose $F$ satisfies the condition. Since $\overline{F^\circ}\subseteq \overline{F} = F$, it suffices to prove $F\subseteq \overline{F^\circ}$.
To this end, fix $x$ in $F$ and let $U$ be an open neighborhood of $x$. Then $x\in U\cap F$ so that $U\cap F\ne\emptyset$. By hypothesis there exists a nonempty open set $U'$ such that $U'\subseteq U\cap F$. As we remarked before, $U'\subseteq F^\circ$ and hence $U\cap F^\circ \supseteq U' \ne \emptyset$. We have shown that every neighborhood of $x$ meets $F^\circ$. Therefore $x\in\overline{F^\circ}$, which proves that $F$ is regular closed. $\square$
Note that none of this requires $X$ to be a metric space.
