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
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)
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' $.