Let $X$ be a topological space and let $A\subseteq X$.
Supose that for each $x\in A$ there exists a neighbourhood of $x$, $V_x$, in $X$ such that $A\cap V_x$ is closed in $V_x$.
Prove that $\Bigl(\bigcup _{x \in A}int(V_x)\Bigr)\cap cl(A)\subseteq A$.
A few notes:
$A\cap V_x$ is closed in $V_x$ $\iff$ There exists a subset $F$ of $X$ which is closed in $X$ and $A\cap V_x=F\cap V_x$, by definition.
$int(A)$ is the interior of A.
$cl(A)$ is the closure of A.
Let $x\in \Bigl(\bigcup _{x \in A}int(V_x)\Bigr)\cap cl(A)$. Please proceed.