closure = union of the set and the set of limit points 
Let $S \subseteq \mathbb{R^n}$ and denote the set of limit points of
  $S$ as $S'$.
Show that $S \cup S' = \bar{S}$ where $\bar{S}$ is the closure.

(i) I want to show that $\bar{S} \subseteq S \cup S'$ . 
Let $x \in \bar{S}$ then either $x \in S$ or $x \in \bar{S}\backslash S $.
If $x \in S$ then $x \in S \cup S'$
If $x \in \bar{S}\backslash S $ ...?
(ii) I want to show that $S \cup S' \subseteq \bar{S}$.
Let $x \in S \cup S'$. Then, $x \in S$ or $x \in S'$.
If $x\in S$ then $x \in \bar{S}$.
If $x \in \bar{S}$ ... ?
Thank you
 A: Looking at (i) you're almost done.  If $x \in \overline{S} \backslash S$ then $x \notin S$ and since $x \in \overline{S}$ then all neighborhoods $U$ of $x$ intersect $S$.  But since $x$ is not in $S$ then $U$ must intersect $S$ at some point other point that is not $x$.  Thus $x \in S'$.  Therefore you can conclude that $\overline{S} \underline{\subset} S \cup S'$.
For (ii) note that if $x \in S \cup S'$ then all neighborhoods $U$ of $x$ intersect $S$ at some point, doesn't matter where.  Thus $x \in \overline{S}$.  Therefore $S \cup S' \underline{\subset} \overline{S}$.
The final conclusion being that $S \cup S' = \overline{S}$
(This is a general result in topology that you can reformulate to real euclidean space if you wish)
A: It is easy to show that $S \cup S' \subset \overline{S}$. Choose $x \in S \cup S'$. If $x \in S$, then clearly $x \in \overline{S}$. So supppose instead that $x \in S'$. Suppose that $x \notin \overline{S}$. Since this is a closed set, its complement is open, and so there is some neighbourhood $U$ of $x$ such that $U \cap \overline{S} = \emptyset$. In particular, $U \cap S = \emptyset$. It should hopefully be not too hard to see that this contradicts the idea of $x$ being a limit point of $S$ (take a sequence $(x_n)$ converging to $x$...).
For the converse, Suppose $x \in \overline{S}$ but that $x \notin S$. Choose balls $U_n$ of radius $1/n$ centered at $x$. Since $x \in \overline{S}$, these will each have non-empty intersection with $S$ (check this!). From here you should be able to build the desired sequence in $S$ which converges to $x$, and so shows that $x$ is a limit point of $S$.
A: Suppose $x\in X$ is a limit point of $A .$ First we note that $\mathrm{int}A \subset A \subset \overline{A} .$ 
Because $A \subset \overline{A}, $ if $x \in A$ then obviously $x \in \overline{A} $. If $ x \notin A ,$ since $x$ is a limit point of $A$, any open $O$ set containing $x$ has the property $ O \cap A \ne \emptyset $ and $  O \cap (X-A)\ne \emptyset $. This means that $x \in \partial A.$ Because $\partial A \subset \overline{A} ,$ we have that $x\in \overline{A}.$
Now suppose $ x \in \overline{A}$. If $x \notin \mathrm{int}A, $ then $x \in \partial A$. Then every open set $O$ containing $x$ has point $y \in A$ and $y\ne x .$ This means $x$ is a limit point.
A: Reply for the comment in the highest score answer:
Suppose $X$ is topology space, $p\in X, A\subset X$, denote the interior, exterior, boundary and closure of $A$ as $A^o, A^e, \partial A$ and $\overline{A}$ respectively. Then $\overline{A}=A^o\cup \partial A= X\backslash A^e$, that is for any nbd. $U$ of $p$, $U\notin X\backslash A\Rightarrow$ $U\cap A\neq \emptyset$.
Definition Used

A: Not sure what you are using as your definition of closed, but one very common definition is: A set S is closed iff it contains all its limit points. If that is a definition you are using or can use, this proof should be easier. 
$\underline{Show: S\cup S' \subseteq \bar S}$: If $x \in S$ then we know $x \in \bar S$ as $S \subseteq \bar S$. If $x \in S'$ then $x \in \bar S$ because $\bar S$ contains all its limit points. Thus $S\cup S' \subseteq \bar S$.
$\underline{Show: \bar S \subseteq S\cup S'}$: Let $x \in \bar S$. We know either $x \in S$ or $x \in \bar S - S$. The first case is trivial, so we will focus on any $x \in \bar S - S$. Since $x\notin S$ but $x \in \bar S$, $x$ must be a limit point of $S$. Thus $x \in S'$. This establishes that for any $x \in \bar S$ either $x\in S$ or $x \in S'$, so $\bar S \subseteq S\cup S' $.
