If A is nowhere dense, then its complement X \ A is dense in X. Let X a topological space. A is nowhere dense in 
 if the interior of the closure of A is empty.
I have to prove that if A is nowhere dense, then its complement X \ A is dense in X.
I tried to prove it,  even looking a similar question but I could not prove it. 
Could anyone give me a valid proof??

If $A$ is nowhere dense, then $int(\bar{A})=\emptyset$. This means that it contains no (non-empty) open sets. Let $U\subset X$ be an open set. Since $A$ is nowhere dense, then $U$ is not a subset of $A$, which means - $U\cap A^c\neq\emptyset$. This means that $A^c$ meets any open set, and it is therefore dense in $X$.
