so to recap: the more modern definition
1) $X$ is a Baire space iff every countable union of closed nowhere dense sets has empty interior.
Equivalent by taking complements (note that a set $A$ is nowhere dense iff its complement $X \setminus A$ contains an open dense subset) to my favourite formulation, which seems to be more commonly used among topologists:
1') $X$ is a Baire space iff every countable intersection of open and dense subsets is dense.
(note that in any space a finite intersection of open and dense subsets is open and dense, so the countable intersection is the first "interesting" question, in a way.)
And what they call the historical definition:
2) Every non-empty open subset of $X$ is of second category.
The article call it historical because it uses a notion "category" of a subset (a subset is either first category or second category, and not both, by definition), which has fallen in some disuse. Nowhere dense sets and meagre sets (the countable unions of nowhere dense subsets) are still normal usage. Note that a first category subset is now called meagre, and the notion of "second category" is not used as much (but still occurs), so it's good to know it. But definition 1) and 2) are easily proved to be equivalent, so they give rise to the same spaces being called Baire. So we have a trivial reformulation of the "classical" definition 2 as:
2') Every non-empty open subset of $X$ is non-meagre.
Or, stated more "positively"
2'') Every meagre set has empty interior.
(otherwise the non-empty interior is a subset of a meagre set, and thus meagre etc.)
which brings us back to definition 1) again.
It's just that the Wikipedians do not like the category terminology (because it might confuse people with category theory as a branch of maths) and so choose to reformulate everything using meagre and non-meagre instead.
Note that the countable closed set (call it $C$, say), also is complete metric or locally compact Hausdorff (whatever $C$ was too). Both properties inherit to closed sets. So Baire category theorem applies to $C$ as well.
Note $x$ is not isolated in $C$ iff $U_x = C\setminus \{x\}$ is dense in $C$. And in both cases $X$ and $C$ have closed singletons, so $U_x$ is open. So if no point is isolated, $$\emptyset = \bigcap_{x \in C} U_x$$ contradicts Baire’s theorem for $C$.
Best Answer
I will instead use the definition that every countable union of closed sets with empty interior has empty interior.
Assume $X$ is scattered, and suppose $\mathcal{A}$ is a countable collection of closed sets with empty interior. Let $U = \bigcup \mathcal{A}$. If $U$ has non-empty interior, then $\text{int}(U)$ contains a point isolated in itself. That is to say, there exists open set $O$ in $X$ such that $\text{int}(U) \cap O =$ {$x$}. Observe that $\text{int}(U)$ and $O$ are both open in $X$, giving us that $x$ is an isolated point in $X$. But $x \in A$ for some $A \in \mathcal{A}$, contradicting $\text{int}(A)$ is empty. Thus, $X$ is a Baire space.