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.
It seems like a more obvious proof is to avoid the corollary and use the theorem directly:
If $x$ is not an isolated point, then $X\setminus\{x\}$ is dense and open in $X$.
If $S=\{s_1,s_2,\dots\}$ is countable, define $V_n=X\setminus \{s_n\}$, which are dense and open since $X$ has no isolated points. Then $\emptyset = \bigcap V_n\cap\bigcap U_n$ is the the countable intersection of dense open sets.
This is essentially the dual of your proof.
Best Answer
The Baire category theorem is an abstraction of the following sort of construction, which seems to arise fairly often. You want to construct an object $X$ (of some given sort) that satisfies an infinite list of requirements, say R1, R2, .... Instead of attacking all these requirements at once, you try to handle them one at a time, and you notice (with considerable happiness) that each requirement is amazingly nice, in the sense that, given any object $Y$, you can slightly modify it to satisfy any single R$n$. So you try starting with some arbitrary $Y$, modifying it to get some $Y'$ that satisfies R1, modifying that to get $Y''$ satisfying R2, and --- OOPS: $Y''$ no longer satisfies R1; the second modification ruined what the first had achieved.
But then you notice an improvement on your original happy observation. The R$n$'s are super-nice in the sense that any $Y$ can be modified to a $Y'$ that satisfies R$n$ and continues to satisfy R$n$ when modified sufficiently slightly.
This means that you can modify $Y$ to some $Y'$ that satisfies R1 and this achievement won't be ruined by future modifications if those modifications are slight enough. Now modify $Y'$ very slightly to get $Y''$ satisfying R2 and still satisfying R1, with room to spare. Here "room to spare" means that further, sufficiently slight modifications won't ruin R1 or R2. Continue in this way, producing $Y'''$ and so forth. After $n$ steps, you have $Y^{(n)}$ satisfying R1 through R$n$ and guaranteed to still do so if perturbed extremely slightly. (At each step, modify the previous $Y^{(n)}$ much less than any of your previous $n-1$ modifications.)
OK, so you can satisfy any finitely many of your requirements with a suitable $Y^{(n)}$, but you want to satisfy all infinitely many requirements with a single object $X$. Well, if you've made the modifications in your construction slight enough, the sequence $Y, Y', Y'',\dots,Y^{(n)},\dots$ will be a Cauchy sequence. If the objects you're working with constitute a complete space, then this sequence will converge to an $X$ that is close enough to each $Y^{(n)}$ to guarantee that $X$ satisfies R$n$. So that limit $X$ is the object you want.
The key ingredients in this argument are (1) completeness of the space; (2) availability of a set $G_n$ of objects that satisfy R$n$ and continue to do so when sufficiently slightly modified, i.e., an open set of objects satisfying R$n$; and (3) the possibility of extremely slightly modifying any object to get it into $G_n$. In other words, $G_n$ is dense in the whole space of your objects.
So, isolating the key ingredients, we have the theorem: In a complete metric space, the intersection of any countably many dense open sets is nonempty.
The complement of a dense open set is a nowhere dense closed set, so the theorem is equivalent to: A complete metric space is not the union of countably many nowhere dense (closed) sets. (I put "closed" in parentheses because omitting it doesn't affect the meaning; a set is nowhere dense iff its closure is nowhere dense.) In other words, a complete metric space is of second category.