B meager, not empty, not open implies complement is dense Let X be topological space. 
The subset B is meager, not open and non-empty. How can you prove that the complement is dense?
 A: This is not true in general. For example if $X=\mathbb Q$ (with the topology inherited from the real line) then every subset of $X$ is countable, and thus meagre. To get a counterexample to your claim, just take an arbitrary subset of $X$, which is not open, and has non-empty interior.  For example, take $B=(-\infty,0]\cap\mathbb Q$.

However, the claim is true if $X$ is a Baire space. Baire spaces are spaces in which Baire Category Theorem holds. For example, every complete metric space and every locally compact Hausdorff space is a Baire space.
Indeed, suppose that $X$ is a Baire space and $B$ is meagre. Then $B=\bigcup_{i=1}^\infty B_i$, where each $A_i$ is nowhere-dense. From this we get $B\subseteq \bigcup_{i=1}^\infty \overline{B_i}$, where $\operatorname{Int\overline{B_i}}=\emptyset$.
Now we get for the complement
$$X\setminus B\supset \bigcap_{i=1}^\infty (X\setminus\overline{B_i}).$$
Each of the sets $X\setminus\overline{B_i}$ is dense in $X$ (since $\operatorname{Int\overline{B_i}}=\emptyset$). In a Baire space, intersection of countably many dense open sets is dense. Therefore $\bigcap\limits_{i=1}^\infty (X\setminus\overline{B_i})$ and, consequently, $X\setminus B$ is dense.
