Prove: A set containing limit points of a sequence is a closed set The question:
Prove that a set, $S'$, containing the limit points of the sequence $S \subset \mathbb{R}$, is closed.
What I have so far:

I want to prove this by showing that the complement of $S'$ is open. In other words, $\mathbb{R} \backslash S' $ is open.
So take any limit point $x$ of $\mathbb{R} \backslash S' $. Clearly, $x$ is not a limit point of $S$.
I'm at a loss on how to proceed from here.
 A: Another nice way to see this would be the following.
Suppose $x\in \bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_n\in S'$, we have the following,
\begin{align*}
x_{11}, &x_{12}, x_{13},\cdots \ \ \ \ \rightarrow x_1  && (\text{a sequence converging to} \ x_1) \\
x_{21}, &x_{22}, x_{23},\cdots \ \ \ \ \rightarrow x_2  && (\text{a sequence converging to} \ x_2) \\
x_{31}, &x_{32}, x_{33},\cdots \ \ \ \ \rightarrow x_3  && (\text{a sequence converging to} \ x_3) \\
\vdots \\
x_{n1}, &x_{n2}, x_{n3},\cdots \ \ \ \ \rightarrow x_n  && (\text{a sequence converging to} \ x_n)
\end{align*}
Note that all the numbers $x_{ij}\in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=\lim_{n\rightarrow\infty}x_n$. This shows that $x\in S'$ by definition of $S'$.  
A: Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(U\setminus\{x\})\cap S=\emptyset$. Now prove $U\cap S'=\emptyset$ and conclude $U\subset(\Bbb R\setminus S')$.
A: If $X$ is any topological space and $S \subset X$, then ${\displaystyle {\bar {S}}=X\backslash [ (X\backslash S)^{\circ  }]}$. Here, we have 
$\tag 1 {\displaystyle {\bar {S}}=\mathbb R\backslash [ (\mathbb R\backslash S)^{\circ  }]}$
Claim: The OP's $S'$ is equal to $\bar {S}$.
It is easy to see that $S' \subset \bar {S}$. 
Using constant sequences, we see that $S \subset S'$. Now let $r \in \bar {S}\backslash S$. By (1), for every $n \ge 1$ the open interval
$\tag 2 (r-\frac{1}{n},r+\frac{1}{n})$
intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r \in S'$. Since both $S$ and its $\bar {S}\backslash S$ are contained in $S'$, $\bar {S} \subset S'$.
Since $S' = \bar {S}$, $S'$ is a closed set.
