Alternative proof of metric space closure for Rudin 2.27 I'm using the definitions from Rudin's Principles of Math. Analysis
Definitions:

*

*If $X$ is a metric space, $E\subset X$, and if $E'$ deonotes the set of all limit points of $E \in X$, the the closure of $E$ is the set $E^{*}= E \cup E'$


*A point $p$ is a limit point of the set $A$ if every neighborhood of p contains a point $q\neq p$ such that $q\in A$


*A set is closed if every limit point of $A$ is a point of $A$
$\ $

Theorem
If $X$ is a metric space, $E\subset X$, then $E^*$ is closed.
Proof. We must show that any limit point $p$ of $E$ lies in $E$. Now any neighborhood $N$ of $p$ contains some $q∈E$. Since $N$ is open, $N$ contains some neighborhood $M$ of $q$, and since $q∈E$, M contains some $r∈E$. Thus $r∈M ⊆ N$, so every neighborhood $N$ of $p$ contains a point $r∈E$, so $p∈E$.

This is not Rudin's proof in the book. I found it on an online errata linked (11th comment on page 3).
Is the following understanding of this of this proof correct? Thanks in advanced.
We must show that any limit point $p$ of $E^*$ lies in $E^*.$ Assume that $p$ is a limit point of $E^*$. Now any neighborhood $N$ of $p$ contains some $q∈E^*.$ Either $q \in E$ or $q\in E'$
If $q\in E$ then by the definition above p is a limit point of $E$ and hence $p\in E'$.
Or $q\in E'$ and there exists some neighborhood $M$ of $q$, such that $M \subset E$ such that it contains a point $r \in E$. It follows that $r \in N \cap E$, therefore $p \in E^*$
 A: "If q∈E
then by the definition above p is a limit point of E and hence p∈E′"
Not quite. $q$ is only one point and we haven't shown every neighborhood contains a point of $E$-- we've only shown that one did.  
Instead the idea is to show that if $q \in E'$ then there is a third point $r$ also in the neighborhood so either $q$ is the point, or there is a third point $r$ in the neighborhood.
.
"Or q∈E′
and there exists some neighborhood M of q, such that M⊂E"
Why is $M \subset E$?  You need to explain why.  Now, I'll confess, I always did a thing where I took a radius that was the $e= \min (d(q,p), radius N- d(q,p))$ and show a neighborhood of $B_e(q) \subset E$.  But I see this proof did something quite a bit simpler and clever.
$N$ is open and $q \in N$ so by the definition of open there is neighborhood $M\subset E$ around $q$.  And being a neighborhood around a limit point $q$ of $E$ there is an $r\in E$ in the neighborhood $M\subset E$ and we are done.
such that it contains a point r∈E. It follows that r∈N∩E, therefore p∈E∗
