If $A$ is locally finite, then so is $\bar{A}$ 


I've read several books and answers to this problem. But what about the case when the open set $W$ doesn't contain any point in $\bar{A}$? (So it misses the boundary points)?
The proposition is basically just $x \in \bar{\Omega} \iff$ every neighborhood contains a point in $\Omega.$
 A: So $W \ni x$ is chosen so that $\{A \in \mathcal{A}: A \cap W \neq \emptyset\}$ is finite. This set can be empty, if $\mathcal{A}$ is not a cover, e.g. but that's no problem, as the empty set is quite finite.
We then just observe that for any open set $W$ and any set $A$ we have $$W \cap A \neq \emptyset \iff W \cap \overline{A} \neq \emptyset\tag{1}$$
The left to right implication is trivial (as $A \subseteq \overline{A}$), and the right to left follows because if some $y \in \overline{A} \cap W$, $W$ is an open neighbourhood of $y$ and $y \in \overline{A}$ then immediately tells us $W \cap A \neq \emptyset$.
So $\{A \in \mathcal{A}: A \cap W \neq \emptyset\} = \{A \in \mathcal{A}: \overline{A} \cap W \neq \emptyset\}$ and $\overline{\mathcal{A}}$ is locally finite. So the case you worry about does not occur.
A: If $W$ is an open subset of $X$ and $b\subset X$ then $$W\cap b\ne\phi \iff W\cap \bar b\ne\phi.$$
Let $A$ be a locally finite family of subsets of $X,$ and let $x\in W\subset X$ where $W$ is open, such that $S= \{a\in A: W\cap a\ne \phi\}$ is finite.
Consider the function $f(b)=\bar b$ for all $b\subset X .$ 
Since $W$ is open, we have $$\{\bar a: a\in A\land W\cap \bar a\ne \phi\}=\{\bar a: a\in A \land W\cap a\ne \phi\}=$$ $$=\{f(a):a\in S\}$$ which is finite because $f$ is a function and $S$ is finite.
