Analysis proof with metric spaces Part a) of Theorem 2.27 in baby Rudin reads (roughly) as follows:

Theorem. Let $X$ be a metric space and $E\subset X$. Then the closure of $E$ is closed.

Proof. Let $x\in\bar{E}^c$. Then $x\notin E$ and $x$ is not a limit point of $E$. Hence $x$ has a neighborhood that does not intersect $E$. The complement of $\bar E$ is therefore open, and hence $\bar E$ is closed. $\square$
The only piece of this proof I don't understand is the conclusion that $\bar E^c$ is open. Haven't we only proved that the neighborhood in question doesn't intersect $E$? Wouldn't it also be necessary to prove that said neighborhood does not contain any limit points of $E$?
 A: If an open set contains a limit point of $E$, then it must also contain a point of $E$ itself (by the definition of limit point), so showing it doesn't intersect $E$ is enough.
A: You are absolutely right.  We need to check that the neighborhood (call it $U$) does not intersect $\overline{E}$.  We already know $U$ doesn't intersect $E$, but it must be shown that $U$ doesn't intersect $\overline{E}$ for $(\overline{E})^{c}$ to be open.
Suppose by contradiction that the neighborhood does intersect $\overline{E}$.  Let $y$ be in $U \cap \overline{E}$.  
Since $U$ is open, there is some $\epsilon>0$ such that $B(y,\epsilon)\subseteq U$.  But $y\in \overline{E}$ implies $B(y, \epsilon) \cap E \neq \emptyset$.  But then since $B(y, \epsilon) \subseteq U$, this implies $U \cap E \neq \emptyset$.  But we said $U$ doesn't intersect $E$, so we've reached a contradiction.
Thus, $U \cap \overline{E} = \emptyset$ which means $U \subseteq (\overline{E})^{c}$, and thus $(\overline{E})^{c}$ is open, as desired.  (And consequently, $\overline{E}$ is closed.)
A: Your goal is to show that the set of limits points for $E$ is closed.
$(1)$ Define $E^* =\{\text{set of limit points of E}\}$
$(2)$ Let $x$ be a limit point of $E^*$.
$(3)$ Then we know any neighborhood of $x$ intersects $E^*$ nontrivially.
$(4)$ Thus, $(N_r(x) -\{x\}) \cap E^*= e_i$ for some $e_i \in E^*$.
$(5)$ But this neighborhood of $x$ is also a neighborhood of $e_1$.
$(6)$ Hence, $x$ is a limit point of $E^*$.
Conclusion: Here we showed that any limit point of $E^*$ is a limit point of $E$ thus contained in $E^* \Rightarrow E^*$ is closed. 
