equivalency of Baire's Theorem I was following Introduction to Topology and Modern analysis by G.F. Simmons. There was,

Theorem D: If $\{A_n\}$ is a sequence of nowhere dense sets in a complete metric space $X$, then there exists a point in $X$ which is not in any of the $A_n$'s.

The proof is straightforward and  there was a equivalent form of Theorem D was mentioned,

Theorem E: If a complete metric space is the union of a sequence of its subsets, then the closure of at least one set in the sequence must have non-empty interior.

Again they mentioned,

Theorem D and E are really one theorem expressed in two different ways. We refer to both (or either) as Baire's theorem.

I don't understand the equivalency of those theorem's and how they are same as Baire's Theorem? How they are related to each other? And from my last year Real Analysis course, I was only introduced this form of Baire's Theorem,

The set of real numbers of $\mathbb R$ can't be written as the countable union of nowhere-dense sets

What will be the general form of this theorem? Because when I googled, I found Baire category theorem only.
 A: I am not familiar with your particular text book, but here is a version of Baire's category theorem for complete metric spaces. From this I will try to show you the equivalences to the results in your posting:

Theorem (Baire): If $X$ is  a complete metric space , then the intersection  of  countable family of open dense sets in $X$  is also dense in $X$.

Proof:*
Let $\{U_n\}$ be a sequence of dense open sets in $X$. Let $B_0$ be a nonempty open set in $X$. We will show that $B_0\cap\bigcap_nU_n\neq\emptyset$.
Given an integer  $n\geq1$, suppose we have chosen a  nonempty set open set $B_{n-1}$. Since $U_n$ is open and dense,  there is a nonempty open set $B_n$ such that $\overline{B_n}\subset U_n\cap B_{n-1}$ (take $B_n$ to be a ball of diameter less than $1/n$). It is clear that
\begin{align}
K:=\bigcap_{n\geq1}\overline{B_n}\subset B_0\cap \bigcap_n U_n.
\end{align}
Observe that the centers of the balls $B_n$ form a Cauchy sequence,
and  by completeness, $K\neq\emptyset$. $\Box$
The result above is  equivalent to the statement

Theorem: In a complete metric space $X$, the union of any sequence of nowhere dense has empty interior.

Proof:
First notice that $E$ is a closed nowhere dense set iff $X\setminus E$ is an open dense set. This follows from
$$ X\setminus\overline{X\setminus E}=\operatorname{int}\big(X\setminus(X\setminus E)\big)=\operatorname{int}(E)$$
Hence, $\{U_n:n\in\mathbb{N}\}$ is a sequence of
open dense subsets of $X$ iff $\{X\setminus U_n:n\in\mathbb{N}\}$ is a sequence of nowhere dense closed sets. Consequently $\bigcap_n U_n$ is dense in $X$ iff $\bigcup_n(X\setminus U_n)$ has empty interior. Therefore, by the Baire category theorem above, in a complete metric space, the union of a sequence $\{E_n:n\in\mathbb{N}\}$ of nowhere dense closed sets has empty interior.
From this, the equivalences follow.
