Property of $A \subset X,$ $X: T_{2}$; subset $A$ of loc. compact $T_{2}$ space $X$ is locally compact iff it is intersection of open and closed set Here's the problem at hand:

a) Let $X$ be a $T_{2}$ space, $A \subseteq X$, $a \in A$, $G \in N(a)$ such that $A \cap G$ is compact. Prove that there exists an open set $V$ such that $a \in V \cap \textrm{cl}(A) \subseteq A$.
b) Let $X$ be a locally compact $T_{2}$ space and let $A$ be its subspace. Show that $A$ is locally compact iff it can be written as the intersection of an open set and a closed set (i.e. iff there exist an open set $V$ and a closed set $F$ such that $A = V \cap F$.

I feel like this problem is a marathon. I've tried a lot, and I haven't even solved a). Here's what I've tried so far.
First, since $G \in N(a)$, there exists an open set $U$ such that $a \in U \subset G$. For every $x \in (A \cap G) \setminus \{a\}$, separate $x$ and $a$ with $x \in U_{x}$, $a \in V_{x}$, $U_{x} \cap V_{x} = \emptyset$. Denote $U_{a} = U$. Then, $A \cap G \subseteq \bigcup_{x \in A \cap G} U_{x}$, so by compactness there exist $x_{1}, ..., x_{n}$ such that $A \cap G \subseteq \bigcup_{i=1}^{n} U_{x_{i}} \cup U$.
Now, if we notice $\bigcap_{i=1}^{n} V_{x_{i}}$, we notice that this is an open set which contains $a$, and is disjoint with $\bigcup_{i=1}^{n} U_{x_{i}}$, so it's pretty close to being the set $V$ that we need; however, I don't know if $\bigcap_{i=1}^{n} V_{x_{i}} \cap \textrm{cl}(A) \subseteq A$ holds. Am I on the right track? How do I proceed from here?
As far as b), I don't really know how to begin with that problem.
 A: For b)
$\Rightarrow)$
Let $F=\text{cl}_X(A)$. We want to prove that there exist an open set $V$ of $X$ such that $A=\text{cl}_X(A)\cap V$, i.e., we want to prove that $A$ is an open set in $\text{cl}_X(A)$.
Let $x\in A$. Because $A$ is locally compact, then there exist $W\subseteq A$ an open set of $A$ such that $x\in W$ and $\text{cl}_A(W)$ is a compact set. Let $W_0\subseteq X$ an open set of $X$ such that $W=W_0\cap A$. We claim that $x\in W_0\cap\text{cl}_X(A)\subseteq A$. We have that $$\text{cl}_A(W)=\text{cl}_X(W)\cap A=\text{cl}_X(W_0\cap A)\cap A$$Because $\text{cl}_A(W)$ is a compact set and $X$ is $T_2$, then $\text{cl}_X(W_0\cap A)\cap A$ is closed in $X$.  In the other hand, $\text{cl}_X(W_0\cap A)=\text{cl}_X(W_0\cap\text{cl}_X(A))$ due to $W_0$ is open. Moreover $$W_0\cap\text{cl}_X(A)\subseteq \text{cl}_X(W_0\cap\text{cl}_X(A))=\text{cl}_X(W_0\cap A) \ \text{and} \ W_0\cap A\subseteq \text{cl}_X(W_0\cap A)\cap A$$Then, $\text{cl}_X(W_0\cap A)\subseteq \text{cl}_X(W_0\cap A)\cap A$ because $\text{cl}_X(W_0\cap A)\cap A$ is closed. Then, $W_0\cap\text{cl}_X(A)\subseteq \text{cl}_X(W_0\cap A)\subseteq A$. Therefore, $A$ is open in $\text{cl}_X(A)$. Thus, there exists an open set $V\subseteq X$ such that $A=\text{cl}_X(A)\cap V=F\cap V$.
$\Leftarrow)$
Because $F$ is closed, then $F$ is locally compact. Moreover, $F\cap V$ is open in $F$ and, as a result of the fact that $F$ is $T_2$, follows that $F\cap V$ is locally compact.
I'm sorry the hard notation. :c
