Is the complement of a dense and open set also dense

Question

I want to show that:
The complement of a dense and open set is nowhere dense.

My Attempt:
Let $A \subseteq X$ be dense, this means $\overline{A}=X$

$\overline{X-A}=X-Int(A)=X-A$

The last equation holds because A is open.

If now $X-A=X \Rightarrow A=\emptyset$
But A being empty is a contradiction to A being dense.

Is this correct?

Answer 1

$\operatorname{Int}(X-A) = X - \overline{A}$ and as $A$ is dense....