A set is closed if and only if it contains all its limit points. 
A set is closed if and only if it contains all its limit points.

Proof in book: Suppose $S$ is "not closed". We must show that $S$ does not contain all its limit points. Since
$S$ is "not closed", $S^c$ is "not open". Therefore there is at least one element $x$ of $S^c$ such that every ball $B(x,\epsilon)$ contains at least one element of $S$ ...
Why is there at least one element $x\in S^c$ such that every open ball contains at least one element of the open set $S$? 
 A: After the edit, everything is fine. Since
$$A \text{ open} \Leftrightarrow A^C \text{ closed}$$
$S^C$ is not open, but an open set is characterised by
$$\forall\ x \in A\ \exists\ \epsilon > 0 \ : \ B_\epsilon (x) \subset A$$
The contraposition is
$$\exists\ x \in A\ : \forall\ \epsilon > 0 : B_\epsilon(x) \cap A^C \neq \emptyset$$
Which is what is stated in the book.
A: $S$ is closed iff $S=\overline{S}$. There is a theorem says $x\in\overline{S}\leftrightarrow\forall U(U\mathrm{\ open}\wedge x\in U\rightarrow U\cap S\neq\emptyset)$. We show $\overline{S}=S\cup S'$ ($S'$ denote the derived set). For $``\supseteq"$, note $S\subseteq\overline{S}$ and if $x\in S'$, then every neighborhood of $x$ intersect $S$ in a point different from $x$, so using the previous theorem, $x\in\overline{S}$. For $"\subseteq"$, suppose $x\in\overline{S}\setminus S$, by the previous theorem, every neighborhood of $x$ intersects $S$, but $x\notin S$ implies it must intersect $S$ in a point different from $x$, so $x\in S'$.
So $S$ closed implies $S'\subseteq S$. and if $S'\subseteq S$, we infer $S=\overline{S}$. QED
Reference: Theorem 17.6, 17.7 of Topology by Munkres, James.
A: Suppose there exists a sequence $x_n$ in $S$ converging to an element $x$ not contained in $S$. Then as a consequence of the definition of convergence, every ball centered at $x$ contains infinitely many elements of $x_n$ (we only need at least 1 for the following argument). Since $x_n$ are not elements of the complement of $S$, this means that the complement of $S$ is not open. This means that $S$ is not closed.
Conversely: Suppose $S$ is not closed.
This means the complement of $S$ is not open.
This means that there exists an $x$ contained in the complement of $S$ such that there doesn't exist a ball centered at $x$ that is contained in the complement of $S$. Hence every ball centered at $x$ intersects with $S$.
Hence there exists a sequence $x_n$ that converges to $x$ defined by taking an element from $S$ contained in the ball centered at $x$ with radius $\frac1n$.
A: For the converse. Assume that S contains its limit points.so if x is in S' then x is in S. Recall that Closure(S)= S U S'. Since Closure of S is the smallest closed set containing S 
i.e S ⊆ Closure(S). Thus, S is closed.
