Locally closed subset equivalence proof using $\bar{L}\cap V = L \cap V$ 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.
 A: The closure operator of $V$ goes as $L\mapsto \bar L\cap V$, no?
Then the property of (a) says that $L\cap V$ is closed in $V$.
A: Let $x\in V\cap \bar L$. We want to show that $x\in V\cap L$. 
We assume that  $x\notin V\cap L$. Since $V\setminus (V\cap L)$ is open in $V$, we have $x\in O\cap V$, where $O$ is open in $E$ and $O\cap V\subset V\setminus (V\cap L)=V\setminus L$. Taking the intersection with $L$ on both sides, we get $O\cap V\cap L\subset V\setminus L \cap L$ hence $O\cap V\cap L$ is empty, which contradicts the fact that $x\in V\cap \bar L$.
The other inclusion is obvious.
A: $\overline{L}$ stands for closure of L below:
To show $\overline{L}\cap V=L\cap V$:
Let $x\in\overline{L}\cap V$. Then $x\in \overline{L}\cap \overline{V}=\overline{L\cap V }=\overline{K\cap V }$ for some closed set $K$ ( as $L\cap V$ is closed in $V$) $=\overline{K}\cap\overline{V}=K\cap\overline{V}$. Hence, as $x\in V$,  $x\in K\cap V = L\cap V$. 
A: We can use the fact that neighborhoods of points in the boundary  $\partial L$ of a subset $L$ of a topological space $X$ have a non-empty intersection with $L$.
If $L \cap V$ is closed in $V$, then $\bar{L} \cap V \subseteq L \cap V$.  
Proof: Let $x\in \bar{L}\cap V$.  Assume $x \notin L$.  Then $x$ must be in $\partial L$.  Since $L \cap V$ is closed in $V$, $V \setminus (V \cap L)$  = $O \cap V$ for some open set $O$ of $X$.  But since $O$ contains $x$ which is in the boundary of $L$, $O \cap L$ is nonempty.  So $O \cap V \cap L$ = $(V \setminus L) \cap L = \emptyset$
