Moreover, why does it follow for Baire Spaces and why is it strictly weaker that:
- $X=\bigcup_{k=1}^\infty A_k\quad\Rightarrow\quad\exists k_0\in\mathbb{N},x_0\in X: A_{k_0}\in\mathcal{N}_{x_0}$
...in words, the space is not meagre iff whenever it is a given by a countable union of subsets then at least one of them is large.
Turning my comment at your other question into an answer
If a Baire space $B$ is the union of countably many sets, then at least one of them has a closure with non- empty interior. But its own interior could still be empty. Take for example $$\Bbb R=\Bbb R∖\Bbb Q∪⋃\{\{q\}∣q∈\Bbb Q\}$$ The set of irrationals as well as each singleton $\{q\}$ has empty interior.