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.
Best Answer
Let $S$ be a set (of real numbers) that does not have the Baire property. Let $\mathcal U$ be the collection of all nonempty open sets $U$ such that $S\cap U$ has the Baire property. Let $\mathcal W$ be a maximal pairwise disjoint subcollection of $\mathcal U$ and let $W=\bigcup\mathcal W$. Since $\mathcal W$ is countable, $S\cap W$ has the Baire property, so the set $X=S\setminus W$ does not have the Baire property.
Since $X$ does not have the Baire property, $X$ is not meager. Let $O$ be a nonempty open set, and assume for a contradiction that $O\setminus X$ is meager.
Since $O\cap W$ is an open subset of the meager set $O\setminus X$, we must have $O\cap W=\emptyset$. Hence $O\cap S=O\cap(S\setminus W)=O\cap X$, which has the Baire property, since $O\setminus X$ is meager. This means that $O\in\mathcal U$, but then the fact that $O\cap W=\emptyset$ contradicts the maximality of $\mathcal W$.