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.
Consider the collection $I$ of all isolated points of $X$. (By the Baire Category Theorem $I$ is nonempty, but that is somewhat immaterial for the moment.) Note that $I$ is then a discrete subspace of $X$. If $I$ were not dense, then $U = X \setminus \overline{I}$ is a nonempty (open) set without isolated points. From here we can construct in the usual manner a Cantor set as a subset of $X$, contradicting that $X$ is countable! (The construction goes as in the linked answer, just ensuring that the $x_\sigma$ are chosen from $U$.)
Best Answer
In fact, being non-meagre (in itself) is a topological property, while being complete (as a metric space) is not. So you have a mismatch: $(0,1)$ is not complete (in its standard metric), but is homeomorphic to $\Bbb R$, which is. As the latter is non-meagre, so is the former. So $(0,1)$ is a trivial example of a non-complete, non-meagre metric space.
More interesting is whether there is a non-completely metrisable space $X$ that is non-meagre. the property of "being completely metrisable" is a topological property. Baire implies that all completely metrisable spaces are non-meagre. But that fails too: $((\Bbb R \times (0,+\infty)) \cup (\Bbb Q \times \{0\})$ (subspace topology from $\Bbb R^2$) is a standard example of a non-meagre metric space that is not completely metrisable.