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.
$\newcommand{\cl}{\operatorname{cl}}$Suppose first that $\langle S,d\rangle$ has the property that the intersection of countably many dense open sets is dense in $S$, and let $A$ be a first category set in $S$; we want to show that $S\setminus A$ is dense in $S$. Since $A$ is first category, there are nowhere dense sets $A_k$ for $k\in\Bbb N$ such that $A=\bigcup_{k\in\Bbb N}A_k$. For $k\in\Bbb N$ let $U_k=S\setminus\cl A_k$; each $U_k$ is a dense open subset of $S$, so $\bigcap_{k\in\Bbb N}U_k$ is dense in $S$. But
$$\bigcap_{k\in\Bbb N}U_k=\bigcap_{k\in\Bbb N}(S\setminus\cl A_k)=S\setminus\bigcup_{k\in\Bbb N}\cl A_k\subseteq S\setminus\bigcup_{k\in\Bbb N}A_k=S\setminus A\;,$$
so $S\setminus A$ is also dense in $S$.
Now suppose that the complement of every first category subset of $S$ is dense in $S$, and let $\{U_k:k\in\Bbb N\}$ be a family of dense open subsets of $S$. For each $k\in\Bbb N$ let $A_k=S\setminus U_k$; $A_k$ is closed and nowhere dense in $S$, so $A=\bigcup_{k\in\Bbb N}A_k$ is first category in $S$. Finally,
$$\bigcap_{k\in\Bbb N}U_k=\bigcap_{k\in\Bbb N}(S\setminus A_k)=S\setminus\bigcup_{k\in\Bbb N}A_k=S\setminus A\;,$$
which is dense in $S$, as desired.
A good book with much information on such topics is John C. Oxtoby, Measure and Category: A Survey of the Analogies between Topological and Measure Spaces, 2nd edition. (The first edition is also good.)
Best Answer
Suppose $Y \subset X$ is of second category in $X$. If possible let $X=\cup_n C_n$ where each $C_n$ is nowhere dense. Then $Y=\cup_n D_n$ where $D_n=Y\cap C_n$. All you have to do is to verify that $D_n$ is nowhere dense. This is obvious because any subset of a nowhere dense set is nowhere dense.
Note: let $X$ be any metric space of first category and $Y$ consist of a single point. Then $Y$ is of second category in itself because it is a complete metric space. So the result is false if the assumption is that $Y$ is of second category in itself. We have to assume that $Y$ is of second category as a subset of $X$.