First of all, sorry for the unspecific title, but can't think of a better one..
Here's an argument.
Let $F$ be a set in a metric space $X$ and $A$ be the set of all isolated points of $F$. Then, $\overline F \setminus A$ is closed and is the set of all limit points of $F$. Since $F\setminus A \subset \overline F \setminus A$, $\overline {F\setminus A} \subset \overline F \setminus A$.
Now, fix $x\in \overline F \setminus A$. Suppose $x\notin \overline {F\setminus A}$. Then there exists $r>0$ such that $B(x,r)\cap (F\setminus A)\setminus \{x\} = \emptyset$. Since $x\notin F\setminus A$, $B(x,r)\cap (F\setminus A) = \emptyset$.
Since $x\in \overline F \setminus A$, $x$ is a limit point of $F$. Thus, $B(x,r)\cap F ≠ \emptyset$. Thus, $B(x,r)\cap F \subset A$ hence $x\in A$, which is a contradiction. Thus $\overline F \setminus A \subset \overline {F\setminus A}$.
Consequently, $\overline F \setminus A = \overline {F\setminus A}$. Q.E.D.
Consider a set $F=\{1/m \in \mathbb{R} | m \in \mathbb{N}\}$. This shows equality does not hold. Where in the argument is wrong and what kind of constraint should I mention?