Why is Baire space itself of 2nd category?
Solution 1:
Yes, straight from that. Suppose $X$ were first category, $X = \bigcup_{n \in \Bbb N} F_n$ where all $F_n$ are nowhere dense.
Then all $\overline{F_n}$ are closed sets with empty interior (by the definition of being nowhere dense) and obviously still $X = \bigcup_{n \in \Bbb N} \overline{F_n}$. But that then contradicts the Baire property of $X$ as $\text{int}(X)=X \neq \emptyset$. So $X$ is second category.