My attempt goes like this:
If $x$ is a limit point of $E$ then every neighborhood of $x$ contains at least one other point of $E$, call it $y$. Since both $x$ and $y$ belong to $E$, they also belong to its closure (call it $\bar E$) and so $x$ is a limit point of $\bar E$.
I want to prove the other side by contradiction. Suppose $x$ is a limit point of $\bar E$ but not $E$. By definition $\bar E = E \cup E'$, where $E'$ is the set of the limit points of $E$. Since $x$ is not a limit point of $E$, $x \notin E'$. Since $E'$ is closed, $x$ is not a limit point of $E'$ either.
This means that there exists at least one neighborhood of $x$ that doesn't contain a point of either $E$ or $E'$. But since $\bar E = E \cup E'$, that same neighborhood doesn't contain a point of $\bar E$ either. Thus $x$ is not a limit point of $\bar E$, a contradiction.
Did I miss some something? The proof in the solutions manual is entirely different, using distances and the definition of neighborhoods and so on.
Best Answer
One problem I see is the statement "both $x$ and $y$ belong to $E$" in your first part of the argument. A priori we do not know that a limit point of $E$ is inside $E$ or not, therefore you cannot say that. What you should say instead is that since $y\in{E}$ and we know $\bar{E}=E\cup{E}'$ so $y\in\bar{E}$, which implies $x$ is a limit point of $\bar{E}$ The next part of the argument is ok, but you can cut it down by just saying that since $\bar{E}$ is closed, it must contain its limit points, but this contradicts $x\notin{E}$ and $x\notin{E}'$. Hope this helps.