$S$ is closed $\iff$ whenever $\{x_n\} \in S$ and $x_n \to x$, then $x \in S$ Definitions:

Limit point: $x$ is a limit point of $S$ if $\forall r>0$, $\exists y \neq x$ such that $y \in B_r(x) \cap S$
Closed$_1$: A set $S$ is closed if its complement is open
Closed$_2$: A set $S$ is closed if and only if it contains all of its limit points

None of these definitions involve sequences, so how can I show both directions when one of the directions involves sequences and the other doesn't?
 A: Hint:
To get one direction, pick a limit point of $S$ and try constructing a sequence that converges to it.
To get the other direction, observe that the limit of a sequence contained in $S$ is a limit point of $S$ and go from there.
A: Assume that $S$ is closed and let $(x_n)_n$ be a sequence in $S$ converging to a point $x$.
Let $\epsilon > 0$. Then, $\exists$ $n_0 \in \Bbb N$ such that: $\forall$ $n \ge n_0$, $d(x_n, x) < \epsilon$
In particular, 
$$d(x_{n_0},x) < \epsilon$$
Thus, $$x_{n_0} \in B_{\epsilon}(x) \cap S$$
So we found that for each $\epsilon > 0$, $B_{\epsilon} \cap S \neq \varnothing $. Thus, $x \in \bar S$. But $S$ is closed, $\therefore x \in S$.
In the other direction, let $x \in \bar S$. For each $n \in \mathbb N$, $\epsilon = \frac{1}{n} > 0$, so we have $B_{\epsilon}(x) \cap S \neq \varnothing$. So, there is a point $x_{\epsilon} = x_{\frac{1}{n}}$ in this intersection.
Define the sequence $(x_n)_n$ as follows: for each $n \in \mathbb N$, pick any point $x_{\frac{1}{n}}$ and let $x_n = x_{\frac{1}{n}}$ (It doesn't matter what the choice of the point is, what matter is that the distance is less than $\frac{1}{n}$).
$(x_n)_n$ is a sequence in $S$. Moreover, since for each $n \in \mathbb N$, $d(x_n, x) < \frac{1}{n}$, we get that $(x_n)_n$ converges to $x$. Therefore, $x \in S$.
But this means that $\bar S \subset S$.
