TITLE: A question about Baby Rudin Theorem 2.27 (a)
QUESTION [12 upvotes]: Theorem 2.27: If $X$ is a metric space and $E \subset X$, then $\bar E$ (the closure of $E$) 
is closed.
The proof says: If $p \in X$ and $p \not \in \bar E$ then $p$ is neither a point of $E$ nor a limit point of $E$. Hence $p$ has a neighborhood which does not intersect $E$. The complement of $\bar E$ is therefore open. Hence $\bar E$ is closed.
I'm particularly questioning about ''Hence $p$ has a neighborhood which does not intersect $E$. The complement of $\bar E$ is therefore open.'' Should we also prove that the neighborhood of $p$ also does not intersect $E'$ (the set of all limit points of $E$)?
Here's what I tried to prove, by contrapositive: ''For any $p \in {\bar E}^c$, if $N_r(p) \cap E' \ne \emptyset$ then $N_r(p) \cap E \ne \emptyset$''.
Proof: For any $p \in {\bar E}^c$, if $N_r(p) \cap E' \ne \emptyset$, then take $q \in N_r(p) \cap E'$, $\exists N_h(q)$ s.t. $N_h(q) \subset N_r(p)$. Since $q \in E'$ is a limit point, $N_h(q) \cap E \ne \emptyset$, and hence $N_r(p) \cap E \ne \emptyset$.
I'm not quite sure whether this is necessary. Or is there anything I missed from Rudin's proof?

REPLY [4 votes]: To amplify slightly on @ncmathsadist's terse comment: It is a matter of negating a definition. If $p$ is not a limit point, then some neighborhood of $p$ must fail to intersect $E$. But then every point $q$ in this neighborhood has that same neighborhood that fails to intersect $E$, so $q\notin \bar E$.
P.S. Rudin is an ambitious choice for self-study. I continue to [complain] that he draws not a single picture, presumably as a matter of pride. Draw lots of pictures!