About the proof of Baire's theorem In my book, the proof of Baire's Theorem starts with:

Let $(X,d)$ be a complete metric space. Suppose, by contradiction, that $X$ is of first category, namely that $X=\bigcup_n C_n \,$ where $\, C_n$ are closed set with empty interior.

On the other hand, the definition of first category should be:

$X$ can be written as the countable union of nowhere dense sets

$\quad$ So:
$$
 X = \bigcup_n A_n \qquad \text{s.t.} \quad \forall n \quad \mathring{\overline{A_n}}= \emptyset 
$$
Therefore, according to the definition, $X$ may be formed by the union of any type of sets (i.e. each $A_n$ can be everything - open, closed, both, neither of them). So why the theorem forces them to be closed?
$\quad$ My understanding attempt is that $\quad \forall n \quad A_n \subseteq \overline{A_n} \quad$ and $\quad \bigcup_n \overline{A_n} \subseteq X$
$\quad$ So that simply $\, \overline{A_n}=C_n \,$ are closed
Am I right?
 A: Yes you are right. If $X=\bigcup_n A_n$ where the $A_n$ are nowhere dense, then by setting $C_n\overset{\text{Def.}}=\overline{A_n}$ you get closed sets $C_n$ with empty interior such that $X=\bigcup_n C_n$.
A: Right. The Baire Category Theorem is that in a complete metric space, if F is 1st-category then the complement of F is dense.  To prove it, it suffices to show that if G is the union of countably many closed nowhere-dense sets then G is co-dense (which means the complement of G is dense)... because any 1st-category F is a subset of, or equal to, some such G.
BTW. A corollary is that if $X$ is a non-empty complete metric space  with no isolated points, and if $Y$ is a dense $G_{\delta} $ subset of $X$ and if $X$  then $Y$ is uncountable:
Let $Y= \cap \{Y_n: n\in \Bbb N\}$ where each $Y_n$ is open. Each $Y_n$ must be dense (because $Y_n\supseteq Y$ and $Y$ is dense.)  Let S be any countable subset of $X,$ with $S\subseteq \{t_n:n\in \Bbb N\}$. (It does not matter whether or not $Y_m=Y_n$ or $t_m=t_n$ for distinct $m,n.$)
Then each $Z_n=^{def}\, Y_n$ \ $\{t_n\}$ is open & is dense (because $Y_n$ is dense and $X$ has no isolated points).  By Baire, $\cap_n Z_n$ is dense in $X,$ and $X$ is not empty so $$Y\setminus S\supseteq \cap_n Z_n\ne \emptyset$$ so $Y\ne S.$
For example, $\Bbb Q$ is dense in $\Bbb R$ but $\Bbb Q$ is only countable so $\Bbb Q$ cannot be a $G_{\delta}$ set in $\Bbb R.$
