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
If you can prove something using choice, then of course you can prove it using Zorn's lemma. Simply prove choice first, then repeat the proof. You can even fold the proofs together and get a slightly more direct proof.
Here is a convoluted outline of such proof. We have our dense open sets, $U_n$, we may assume they are decreasing in inclusion, of course. To show the intersection is dense, pick a non-empty open set $W$ and define a sequence of points $x_n\in U_n$, such that the sequence converges to a point in $W$.
To use Zorn's lemma here, consider the partial order of possible finite sequences for this proof. If every such sequence is finite, then every chain is bounded from above (trivially) and there is a maximal element. But an easy argument shows that no maximal element can exist, because of density and openness of the $U_n$'s. Therefore there is a chain without an upper bound, which must mean it is infinite. And this provides the sequence we want.
The lesson to learn here is that choice has many variants and equivalents and weak forms. And using one kind of equivalent might be simpler in some proofs than another.