Proving the Baire Category Theorem from scratch, Stuck! 
Theorem: (Baire) 
$(X,d)$ is a complete metric space then the intersection of countably
  many dense, open subsets in the metric topology $\mathcal{T}$ generated by $d$ is
  dense

In other words, 

$$D = \bigcap_{n \in \mathbb{N}} D_n$$ where $D_n$ is a dense, open
  subset of $X$, is dense

I am asked to prove this raw, no hints whatsoever. I am already hitting my head on several bricks

Idea/Approach: By contradiction, assume that $D$ defined above is not dense, then $(X,d)$ is not complete. 
Proof Attempt: (No idea how to start! We will start by stating what we want)


*

*We wish to construct a Cauchy sequence $(x_n)$ on $X$, then by completeness of $X$, $x_n
   \to x$ as $n \to \infty$. Afterwards, we wish to obtain a contradiction such that if $D =
   \bigcap_{n \in \mathbb{N}} D_n$ is not dense, then $x_n \not\to x$.


(Difficult part is to find out how $D_n$ is related to $x_n$...)


*

*Let $\{D_n\}$ be a set of dense, open subset of $X$, where $n \in \mathbb{N}$. By definition, for all $U \in \mathcal{T}$, $D_n \cap U \neq \varnothing$.

*Let $D =   \bigcap_{n \in \mathbb{N}} D_n$ and suppose for contradiction there exists $U \in \mathcal{T}$ such that $D \cap U = \varnothing$ (i.e. $D$ is not dense). Then let $x_n \in D_n \cap U, \forall n \in \mathbb{N}$

*Then $(x_n)$ is a Cauchy sequence if $\forall \epsilon > 0, \exists N \in \mathbb{N}$, such that $d(x_n, x_m) < \epsilon, \forall n, m > N$
(At this point it is obvious that without additional assumptions, there is no way $(x_n)$ is Cauchy)
... 
Does anyone see if this approach can still be continued? Any help is appreciated at this point.
 A: Your approach can be carried out; you just have to figure out how to choose the points $x_n$ so that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence. I’ll get you started.
For $x\in X$ and $r>0$ let $C(x,r)=\{y\in X:d(x,y)\le r\}$, the closed ball of radius $r$ centred at $x$. 
Begin by choosing any $x_0\in U\cap D_0$; there is an $r_0>0$ such that $C(x_0,r_0)\subseteq U\cap D_0$. The open ball $B(x_0,r_0)$ intersects $D_1$, so choose $x_1\in B(x_0,r_0)\cap D_1$. There is an $r_1>0$ such that $C(x_1,r_1)\subseteq B(x_0,r_0)\cap D_1$ and $r_1\le\frac12r_0$. (Why?) Similarly, $B(x_1,r_1)\cap D_2\ne\varnothing$, so there is an $x_2\in B(x_1,r_1)\cap D_2$, and there is then an $r_2>0$ such that $C(x_2,r_2)\subseteq B(x_1,r_1)\cap D_2$ and $r_2\le\frac12r_1$. (Again, why?) Now try to complete the blockquoted sentence below that describes how to keep going in this fashion.

Given $x_n\in U$ and $r_n>0$, we know that $B(x_n,r_n)\cap D_{n+1}\ne\varnothing$, so there is an $x_{n+1}\in B(x_n,r_n)\cap D_{n+1}$, and there is then an $r_{n+1}>0$ such that ...

In the end we have a sequence $\langle x_n:n\in\Bbb N\rangle$ such that each $x_n\in U$.


*

*Show by induction on $n$ that $x_n\in\bigcap_{k\le n}B(x_k,r_k)$ for each $n\in\Bbb N$.  

*Use this to show that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence and therefore converges to some $x\in X$.  

*Show that for each $n\in\Bbb N$, $x_n\in\bigcap_{k\le n}D_k$.  

*Use this to show that $x\in\bigcap_{n\in\Bbb N}D_n$.


Note that a proof by contradiction is not needed: the argument shows directly that $\bigcap_{n\in\Bbb N}D_n$ meets every non-empty open $U\subseteq X$.
