Theorem A set E is open if and only if the complement $E^{c}$ is closed
Proof
$E$ open $\iff$ any point $x \in E$ is an interior point
$\iff \forall x \in E, \exists$ a neighborhood $N$ of x s.t. $N$ is disjoint from $E^{c}$
$\iff \forall x\in E$, $x$ is not a limit point of $E^{c}$
$\iff E^{c}$ contains all its limit point
Here is my question for the last justification. How does showing that $x$ is not a limit point of $E^{c}$ imply that the $E^c$ contains all the limit points?