Let me give you some extra tools to use, here. Some details are left to you to verify.
Definitions: A subset $A$ of a topological space $X$ is said to be nowhere-dense (in $X$) if for every non-empty open subset $U$ of $X$ there is a non-empty open subset $V$ of $X$ such that $V\subseteq U$ and $V\cap A=\emptyset$. (You should be able to prove that this is equivalent to your definition of nowhere-dense.) A topological space $X$ is said to be a Baire space if every countable union of nowhere-dense subsets of $X$ has empty interior.
Lemma 1: $X$ is a Baire space if and only if every countable intersection of open dense subsets of $X$ is dense in $X$.
Proof: Suppose $X$ is a Baire space and $\{G_n\}_{n=1}^\infty$ is a countable collection of open dense sets, so each $X\setminus G_n$ is nowhere dense. (Why?) Thus, for any non-empty open set $U$, we have that $$U\nsubseteq\bigcup_{n=1}^\infty (X\setminus G_n)=X\setminus\left(\bigcap_{n=1}^\infty G_n\right),$$ so there must be some point of $\bigcap_{n=1}^\infty G_n$ in $U$. Thus, $\bigcap_{n=1}^\infty G_n$ is dense.
Suppose countable unions of open dense sets are dense, and let $A_n$ be nowhere dense for each $n$, so each $X\setminus\overline{A_n}$ is open and dense. (Why?) Take any non-empty open $U$, so that $U$ intersects $$\bigcap_{n=1}^\infty\left(X\setminus\overline{A_n}\right)=X\setminus\left(\bigcup_{n=1}^\infty\overline{A_n}\right)\subseteq X\setminus\left(\bigcup_{n=1}^\infty A_n\right),$$ so cannot be a subset of $\bigcup_{n=1}^\infty A_n$. Thus, $X$ is a Baire space. $\Box$
Lemma 2: $X$ is a Baire space if and only if every countable union of closed nowhere-dense subsets of $X$ has empty interior. (I leave the proof to you. Lemma 1 and DeMorgan's Laws should help.)
Lemma 3: If $X$ is a non-empty complete metric space with metric $d:X\times X\to\Bbb R$, and $F$ is a non-empty closed subset of $X$, then $F$ is a complete metric space with metric $d_F$ defined by $d_F(x,y)=d(x,y)$ for all $x,y\in F$. (I leave the proof of this to you.)
Proposition: Assuming BCT holds, every non-empty complete metric space is a Baire space.
Proof: Suppose by way of contradiction that $X$ is a non-empty complete metric space, but isn't a Baire space. Then by Lemma 2, there is some countable collection $\{A_n\}_{n=1}^\infty$ of closed nowhere-dense subsets of $X$ whose union doesn't have empty interior--meaning there is some non-empty open $U$ such that $$U\subseteq\bigcup_{n=1}^\infty A_n.$$ Now, in particular, we can take a non-empty open set $V$ such that $\overline V\subseteq U.$ (Why?) Then $$\overline V\subseteq\bigcup_{n=1}^\infty A_n,$$ so $$\overline V=\overline V\cap\left(\bigcup_{n=1}^\infty A_n\right)=\bigcup_{n=1}^\infty(\overline V\cap A_n).\tag{#}$$ But $\overline V$ is a non-empty complete metric space by Lemma 3, so by BCT, $\overline V$ is not a countable union of closed nowhere-dense subsets of $\overline V$. But each $\overline V\cap A_n$ is closed and nowhere-dense in $\overline V$ (why?), so $(\#)$ gives us the desired contradiction. $\Box$
The term meagre is the one that provides the basic intuition. A nowhere dense set is clearly small in a topologically meaningful sense: it’s so small that its closure has empty interior. A meagre set is small in the sense that it can be expressed as the union of just countably many nowhere dense sets. That’s not quite so small as being nowhere dense, since a meagre set can be dense in the whole space (e.g., $\Bbb Q$ in $\Bbb R$), but this is still clearly a notion of smallness. And it’s a nicer one to work with than the basic notion of nowhere dense set, because the meagre sets form a $\sigma$-ideal: all subsets and countable unions of meagre sets are still meagre. (The nowhere dense sets merely form an ideal: all subsets and finite unions of nowhere dense sets are nowhere dense.)
The Baire category theorem then says that if a space is nice enough (complete metric space, locally compact Hausdorff space, etc.), then it isn’t small (i.e., meagre).
More generally, any ideal or $\sigma$-ideal $\mathscr{I}$ of subsets of some set $X$ can be considered a notion of smallness for subsets of $X$. Sets in $\mathscr{I}$ are small; complements of sets in $\mathscr{I}$ are large; and all other subsets are neither small nor large. Some of these notions of smallness turn out to be very useful, and the $\sigma$-ideal of meagre sets of a topological space is one of the useful ones. The ideal of non-stationary subsets of a cardinal of uncountable cofinality is another one that turns out to be very useful.
Best Answer
Assume $A$ is of first category, and assume, toward a contradiction, that its complement $A^c$ is also of first category. So $A$ is the union of countably many nowhere dense sets $A_n$ and $A^c$ is the union of countably many nowhere dense sets $B_n$. By definition of "nowhere dense", the closures $\overline{A_n}$ and $\overline{B_n}$ have empty interiors. But the union of all these closures inlcudes both $A$ and $A^c$, so it is the whole space, and that contradicts Baire's theorem.