If a subset of a top. space is closed, then it contains all its limit points.

So I need to show that if $\displaystyle A\subset X$ is closed then $\displaystyle A=\bar{A}$.

Since A is closed, $\displaystyle X-A$ is open. Also, $\displaystyle \bar{A}=A\cup A'$

A' is the set of all limit points of A.

I am not sure if this helps, but

$\displaystyle X-\bar{A}=X-(A\cup A')=(X-A)\cup (X-A')$