I would like general feedback on my solution to this exercise, as well as any ideas for a more direct approach.
Exercise
Let $(X,d)$ be a metric space. Denote the closure of any set $A\subseteq X$ by $\bar{A}$.
Prove that $\bar{\bar{A}}=\bar{A}$.
Solution
Clearly $\bar{A}\subseteq\bar{\bar{A}}$, so it suffices to prove that any limit point of $\bar{A}$ is also a limit point of $A$.
Let $x\in X$ be a given limit point of $\bar{A}$, and let $\varepsilon >0$ be given.
If $x\in A$ then clearly $x\in\bar{A}$ so for the remainder of the proof suppose $x\notin A$.
So $\exists y\in\bar{A}$ such that $x\neq y$ and $d(x,y)<\frac{\varepsilon}{2}$.
There are two possibilities:
Case 1 ($y\in A$)
In this case, $y\in A$ is such that $x\neq y$ and $d(x,y)<\frac{\varepsilon}{2}<\varepsilon$.
So $x$ is a limit point of $A$, as required.
Case 2 ($y$ is a limit point of A)
In this case, $\exists z\in A$ such that $z\neq y$ and $d(z,y)<\frac{\varepsilon}{2}$.
So $z\in A$ is such that $z\neq x$ (since $x\notin A$) and $d(x,z)\leq d(x,y)+d(y,z)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$.
So $x$ is a limit point of $A$, as required.
