Show that if every nonempty open set in $X$ is non-meager, then every comeager set in $X$ is dense Suppose $X$ is a topological space. Prove that if every nonempty open set in $X$ is non-meager, then every comeager set in $X$ is dense.
My attempt:
Suppose $A \subset X$ is a comeager set and let $O \subset X$ be an open set. I want to show that $A \cap O \neq \emptyset$. 
By definition of comeager set, $A$ contains $\cap_{n \in \mathbb{N}}{A_n}$ where $A_n$ is dense and open for all $n \in \mathbb{N}$. 
By definition of denseness, $A_n \cap O \neq \emptyset$ for all $n \in \mathbb{N}$. Hence, $(\cap_{n \in \mathbb{N}}{A_n}) \cap O \neq \emptyset$. Since $\cap_{n \in \mathbb{N}}{A_n} \subset A$, we have $(\cap_{n \in \mathbb{N}}{A_n}) \cap O \subset A \cap O \neq \emptyset$.
Question: 
Is my proof correct? I don't think so. Because the assumption is not used anywhere in the proof. 

No.  Although, for all $n$, the set $A_n\cap O$ is not empty, it might contain entirely different points for different $n$.  So we may not conclude $\bigcap_n{A_n}\cap O$ nonempty.  This is precisely why you need the unused hypothesis.  
On the other hand, note that if $$\emptyset=\bigcap_n{A_n}\cap O$$ then, taking the complement in $O$, we should have $$O=\bigcup_n{(O\setminus A_n)}$$  But $A_n$ is dense and open in $O$.  What does this say about $O\setminus A_n$ (and, by extension, $O$)?
