Properties of Closure Definitions:

$A'$ is the set of all accumulation or limit points.

$\bar{A} = A \cup A'$ - this is known as the closure of $A$.

Let $A$ be a subset of $\mathbb{R}$. A point $p\in\mathbb{R}$ is an accumulation point of $A$ if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$.

Prove or disprove: $(\overline{A})^{c} = \overline{(A^{c}})$

proof: This is true since we know what the compliment of the closure set is open. 

I am not sure if I am right, assume already that I have proven that the compliment of the closure set is open.
The statement is not true. If $A$ is the set of rationals, then $(\bar{A})^c = (\Bbb R)^c = \emptyset$, but $\overline{A^c} = \bar{\Bbb I} = \Bbb R$, where $\Bbb I$ is the set of irrationals.
This is not generally true. Furthermore, we can make the following claim: 

Let $X$ be a connected topological space and $A\subseteq X, A\neq\emptyset, A\neq X$. Then $\left(\overline{A} \right)^c\neq\overline{A^c}$.

Assume the contrary. Then $\left(\overline{A}\right)^c$ is open, but since $\overline{A}=\left(\overline{A}\right)^{c^c}=\left(\overline{A^c}\right)^c$, we also have that $\overline{A}$ is open. Then $X$ is the union of two disjoint nonempty open sets, which contradicts $X$ being connected.