Proof involving Closure of Sets Given definitions: 
$ S\subseteq \Bbb R^n , S \neq \emptyset  $
The closure of $S$, denoted cl($S$), is defined as $S$  $\cup$ $S'$ where $S'$ is the set of all limit points or accumulation points of $S.$
I must show that $S'$ = cl($S'$)
Below is my proof, I'd appreciate any corrections and advice.
Proof:
In order to show $S'$ = cl($S'$), I must show double inclusion, that is $S'$ $\subseteq$ cl($S'$)  and cl($S'$) $\subseteq$ $S'$ 
i) $S'$ $\subseteq$  cl($S'$) because for $\forall$ $x_i \in S'$ for $i \in \Bbb N$, $x_i \in$ cl($S'$) since cl($S$) = $S \cup S'$

ii) cl($S'$) $\subseteq$ $S'$ since cl($S'$) = $S' \cup S''$ where $S''$ is the set of all limit points of $S'.$
I'm looking to see if this is makes sense mathematically. Often I have a hard time putting into words what make sense to me in my head! Also, if this proof is accurate mathematically, I would like to know if it is expressed in a clear manner. 
Thanks for your insight! 
 A: A point  $p$  is a limit point of $S$ iff every neighborhood $V$ of $p$ contains a point $q \in (S \cap V)$ with $q \ne p$. For part (ii) you need to show $S'' \subset S'$............. Let $p \in S''$............ Let $V$ be any nbhd of $p$. There exists $q \in V\cap S'$ with $q \ne p$. Now $q$ has a nbhd $W$ with $W\subset V$ and $p\not \in W$,and since $q \in S'$, there exists $r\in W\cap S$ . Now we have $r \in S\cap V$ (because $r \in W \subset V$) and $r \not = p$ (because $r \in W$ and $p \not \in W$. So every nbhd $ V$ of $p$ has a point $r$ in $S\cap V $ with$ r \not = p$.............. So $p \in S'$.  
A: Since $cl(S')=S'\cup S''\supset S'$, i) is obvious. 
For ii), we prove that $S''\subset S'$.
Let $x\in S''$. If $x\in S'$, then done. If not, by the definition of accumulation points, for any neighborhood $G_1$ (open set) of $x$, there is $G_1\cap (S'-\{x\})\ne \varnothing$. Pick a $y\in G_1\cap (S'-\{x\})$ and $y\ne x$. 
If $y\in S$, then neighborhood of $x$ contains a point of $S$ and so $x$ is an accumulation point of $S$, or $x\in S'$, done. If not,  then there is a neighborhood $G_2$ of $y$ that  $G_2\subset G_1, \:G_2\cap (S-\{y\})\ne \varnothing$ for if $G_2$ is not a subset of $G_1$, choose $G_2=G_2\cap G_1$ instead. Pick a $z\in G_2\cap (S-\{y\})$. Then $z\in G_1$ and  $z\in S$, which means that $x$ is an accumulation point of $S$ or $x\in S'$. So $S''\subset S'$. 
