# Prop: $G \subseteq X$ is open in $X$ $\implies$ $cl(G\bigcap cl(A)) = cl(G\bigcap A)$ for every $A \subseteq X$.

Denote the closure operator of a topological space by $$cl$$.

Prop: $$G \subseteq X$$ is open in $$X$$ $$\implies$$ $$cl(G\bigcap cl(A)) = cl(G\bigcap A)$$ for every $$A \subseteq X$$.

Pf: Assume that $$G$$ is open in $$X$$ and $$A \subseteq X$$. It is trivial that $$G \bigcap A \subseteq G \bigcap cl(A) \implies cl(G\bigcap A)\subseteq cl(G\bigcap cl(A))$$. Further assume that $$x \in cl(G\bigcap cl(A))$$. Then for every $$x \in U \subseteq X$$, $$U\bigcap (G\bigcap cl(A))\neq \emptyset$$. Using the fact that $$G$$ is open, there exists $$y \in U$$ such that for $$y \in V \subseteq X$$, $$V \subseteq G$$. Since $$(V\bigcap G)\bigcap cl(A) \neq \emptyset \implies V \bigcap cl(A)\neq \emptyset \implies V \bigcap A \neq \emptyset$$, $$U \bigcap (G\bigcap A)\neq \emptyset$$ and $$x \in cl(G \bigcap A)$$ as required.

Does my proof look correct?

Prove the following highly useful lemma:
open U implies U $$\cap$$ cl K subset cl (U $$\cap$$ K).

Thus cl (U $$\cap$$ cl K ) subset
cl cl (U $$\cap$$ cl K) = cl (U $$\cap$$ K)
subset cl (U $$\cap$$ K).

Your proof fails because U and V are just any sets.