Prove that a set with disjont neightborhodd of a boundery and interior is nowhere dense

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I suck in this problem, suppose $F$ is closed, $G$ is a open neighborhood of $x\in F$ while $G \cap F^{\circ}=\emptyset$, is it enough to show that $F^{\circ}=\emptyset$? If not, how about that in a metric space? or even in $\mathbb{R}$.

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$F=\{1\}\cup [2,3], x=0, G=(-1,1)$ is a counter-example.