Let $X$ be a compact Haussdorf topological space . Prove that if $\lbrace A_{n}: n \in \mathbb{N} \rbrace$ is a sequence of nowhere dense sets of $X$. Then $X-\bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace$ is dense in $X$.
Since $\lbrace A_{n}: n \in \mathbb{N} \rbrace$ is a sequence of nowhere dense sets of $X$, we have that $int\overline{(A_{n})} = \emptyset$ for every $n \in \mathbb{N}$. So what I intend to prove is that $ \overline{X-\bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace}=X$. Obviously since $X-\bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace \subset X$, we have that $\overline{X-\bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace} \subset \overline{X}=X$, the last contention because $X$ is clopen. So I only need to prove the contention $X \subset \overline{X-\bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace}$. By definition of complement of a closure $\overline{X-\bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace}= X-int(\bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace)$. Here I try to proceed by contradiction, asume $x \in X$ and $x \in int(\bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace)$, this means there is an open set $U$ such that $x \in U$ and $U \subset \bigcup \lbrace A_{n}: n \in \mathbb{N} \rbrace$. If countable union of nowhere dense sets is a nowhere dense set we are done. But this is not always true. However, I was wondering how to solve this whether with my idea or another aproach? Still dont know how I can use the hypothesis of $X$ being compact or Haussdorf.