From 'Treatise on Analysis (vol 2)':
(12.2.3) Let $L$ be a subset of a topological space $E$. Then the following properties are equivalent: (a) For each $x\in L$ there exists a neighborhood $V$ of $x$ in $E$ such that $L\cap V$ is closed in $V$;
(b) $L$ is an open subset of the subspace $\bar{L}$ (the closure of $L$ in $E$);
(c) $L$ is the intersection of an open subset and a closed subset of $E$.
The book proves this with b $\Rightarrow$ c, c $\Rightarrow$ a, a $\Rightarrow$ b. The last one however (a to b) is what I'm stuck on. I am following their approach exactly for this particular proposition.
Their proof for a $\Rightarrow$ b goes:
For each $x \in L$, we have $V \cap L = V \cap \bar{L}$, because $V \cap L$ is closed in $V$; this shows that in the subspace $\bar{L}$ the point $x$ is an interior point of $L$, and therefore $L$ is open in $\bar{L}$.
So the part I'm stuck on is showing that $\bar{L} \cap V = L \cap V$. I wasn't sure whether the closure in that expression was w.r.t. subspace $V$ or the space $E$. But assuming either leaves me stuck.
Hints are more welcome, so that some work is left for me to learn from.
Grazie.