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.
Let $A$ be a nowhere dense set. Then $(\overline{A})^o=\emptyset$.
This is equivalent to saying that $(\overline{A})^c$ is dense in $X$.
Let $A$ and $B$ be two nowhere dense sets. Let $S=A \cup B$.
To show $S$ is nowhere dense we will show that $(\overline{S})^c$ is dense in $X$, that is $(\overline{S})^c$ meets every non-empty open set.
Let $G$ be a non-empty open set.
Now as $A , B$ are nowhere dense, $G\cap(\overline{A})^c\neq\emptyset$ and $G\cap(\overline{B})^c\neq\emptyset$.
Also $(\overline{A})^c$,$(\overline{B})^c$ are open.
Hence $G\cap(\overline{A})^c\cap(\overline{B})^c\neq\emptyset$, since $G\cap(\overline{A})^c$ is non-empty open and $(\overline{B})^c$ is dense in $X$.
Thus $G\cap(\overline{A}\cup\overline{B})^c\neq\emptyset$, which implies $G\cap(\overline{S})^c\neq\emptyset$.
Hence $(\overline{S})^c$ is dense in $X$. Equivalently $S$ is nowhere dense.
Best Answer
The second one is correct. To show that a space is of second category you have to show that if the space is expressed as a countable union of sets then the interior of closure of at least one of then is non-empty. The first one is false.