A proof: Closure of a set of complex numbers is closed I'm assuming the proof should apply for complex, and real numbers. I've offered a proof for complex numbers, and I am not so sure about some of the steps (namely $a_n \in \bar S_c)$ I take so a second look/hints would be appreciated! The only other theorem I have is that a closed set implies every sequence in the set converges to a point in the set. 
Let $\bar S = S \cup \partial S$ denote the closure of S, with $\partial S$ being the boundary. Also let $\bar S_c$ be the complement of the closure (sorry if that is unconventional). 
Suppose $\bar S_c$ is closed. Consider $z \in \partial S$, then $D(z, r) \cap S_c \neq \emptyset$ for any $r > 0$. Take $a_n \in D\left(z, \frac{r}{n}\right) \cap S_c$ for $n = 1, 2, ...$ Then $a_n \in \bar S_c$ because $a_n \notin S, a_n \notin \partial S$. Clearly, $a_n \rightarrow z$ as $ n \rightarrow \infty$. However, $z \in \bar S$ and $a_n \in \bar S_c$ which means there is a convergent sequence in $\bar S_c$ converging to a point outside $\bar S_c \implies \bar S_c$ is open $\implies \bar S$ is closed.
 A: You can prove your result directly from the definition of closure.
And not that $\bar{S}$ contains all the limit points of $S$
Let $x \notin \bar{S}$
Then exists $r>0$ such that $B(x,r) \cap S=\emptyset$ and $B(x,r)$ does not contain any limit point of $S$ because if it had a limit point $y$ of $S$ then  from the definition of the limit point $B(x,r)$ as a neighborhood of $y$ would intersect $S$ which is absurd.
Thus $B(x,r) \subseteq \mathbb{C}$ \ $\bar{S}$.
So for every element $x$ in the complement of the closure of $S$ we find an open ball with center $x$ which is a subset of the complement.
So  $\mathbb{C}$ \ $\bar{S}$ is open.
Also $a_n$  need not to belong to  $\bar{S}^c$
Take for instance the interval $[0,1]$.
We know that $1 \in \partial{([0,1])}$
Take $$a_n=\begin{cases}\
          1-\frac{1}{n} & n=2k+1\\
          1+\frac{1}{n} & n=2k\\
          \end{cases}$$
and $a_n \to 1$
A: We will prove that $$\bar S =\bigcap_{\ \ \ S\subseteq F\\ F\ \text{closed}} F.$$
Not only will this give us that $\bar S$ is closed as intersection of closed sets, it will give us extremely useful characterization of closure (which I personally prefer to be definition): the closure of $S$ is the smallest closed set containing $S$.
Let $F$ be closed such that $S\subseteq F$. We want to prove that $\bar S = S\cup \partial S\subseteq F$ and by the choice of $F$, it is enough to prove that $\partial S\subseteq F$. 
Assume the contrary, i.e. there exists $x\in\partial S$ such that $x\not\in F.$ Then, $x\in F^c$, which is open. Thus, there exists open ball $B(x,r)$ such that $B(x,r)\subseteq F^c$. But, since $x\in\partial S$, it means that $B(x,r)\cap S\neq\emptyset$. Let $y\in B(x,r)\cap S$. In particular, $y\in S\subseteq F$, so $B(x,r)\cap F\neq\emptyset$. But this is in contradiction with $B(x,r)\subseteq F^c$. Thus, $\partial S\subseteq F$ and consequently $\bar S\subseteq F$.
Since $F$ was arbitrary, it follows that 
$$\bar S \subseteq\bigcap_{\ \ \ S\subseteq F\\ F\ \text{closed}} F.$$
Let us now prove the other inclusion. Let $$x\in\bigcap_{\ \ \ S\subseteq F\\ F\ \text{closed}} F$$ and assume that $x\not\in S$. We will prove that $x\in\partial S$ then, which will complete our proof.
If $x\not\in\partial S$, then there exists open ball $B(x,r)$ such that $B(x,r)\subseteq S^c$. Let $F = B(x,r)^c$. Then, $F$ is closed and $S\subseteq F$, so by our choice of $x$, $x\in F$. Contradiction. Thus, if $x\not\in S$, then $x\in\partial S$, which proves $x\in S\cup\partial S$.
A: A set is closed if and only if it contains its limit points.
$\partial S = \{\text{limit points  of } S\} \cap \{\text{limit points of } S^c \}$.
It suffices to prove:
Lemma:  If $y$ is a limit point of $\overline S = S\cup \partial S$ then $y$ is a limit point of $S$.
Thus every limit point $\overline S$ is in $\{\text{limit points  of } S\} \subset \overline S$.  So $\overline S$ is closed.
Proof of Lemma:
Suppose not.  Let $y$ be a limit point of $\overline S$ that is not a liit point of $S$.  Then there exist an $\epsilon > 0$ so that $B_{\epsilon}(y)$ contains no points (other than possible $y$ itself) that are in $S$.
But $y$ is a limit point of $\overline S$ so $B_{\epsilon}(y)$ contains a $w \in \overline S$.  But $w \not \in S$ so $w \in \overline S \setminus S = \partial S \subset \{\text{limit points of } S\}$.
So $w$ is a limit point of $S$.  Let $\delta = \min(d(w,y), \epsilon - d(w,y))$.
The there exists a point $z\in S$ that is in $B_{\delta}(w)$.
But $d(z,y) \le d(w,z) + d(w,y) < \delta + d(w,y) \le \epsilon - d(w,y) + d(w,y) = \epsilon$.  So $z \in B_{\epsilon} (y)$.
Which is a contradiction.
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One graphical way of viewing limit  points is $z$ is a limit point of $X$ if $\inf \{d(x,z)|z \in X\} = 0$.
Which would mean if $x$ is a limit point of $\overline S$ means $\inf d(x,y):y\in \overline S = 0$ but $\inf d(y, s):s \in S = 0$ .  So $\inf d(x,s)= 0$.  So $x$ is a limit point of $S$.
