Proof attempt of Baire's Category theorem. Usually I do my pure mathematics proofs on my own.Now I am studying Baire's Category theorem which states that a complete metric space cannot be meagre(i.e.it is of second category).I tried to prove it but I am stuck at a particular point.The proof of mine is given below:
Suppose $X$ is meagre.Then $X=\large\cup_{n\in \mathbb N}A_n$ where each $A_n$ is nowhere dense.Now without loss of generality we can replace $A_n$ by $\bar A_n$ in other words, we can assume $A_n$ to closed.But notice that each of them has an empty interior.Notice that each $A_n$ is a proper subset of $X$,so that $X-A_n$ is not empty.We shall prove that $X$ is an incomplete metric space.Notice that $\large\cap_{n\in \mathbb N}(X-A_n)=\phi$.Where each $(X-A_n)$ is open and dense in $X$.Define $B_n:=X-A_n$.Now $B_n$ is non-empty and open so contains an open ball of radius $r_1$ namely $B(x_1,r_1)$,where $r_1<1/2$ and $B_2$ is dense.So,$\exists x_2 \in B_2\cap B_{r_1}(x_1)$,now this set being open also contains $B_{r_2}(x_2)$ such that $r_2<1/2^2$,now $\exists x_3 \in B_3\cap B_(x_1,r_1)\cap B(x_2,r_2)$,and so we continue this process to get hold of a sequence $(x_n)$ such that $d(x_n,x_{n-1})<1/2^n$,so $d(x_m,x_n)\leq \sum_{k=m}^n 1/2^k$ and thus as $m,n\to \infty$,$d(x_m,x_n)\to 0$.So this sequence is Cauchy,Now I want to prove that this does not converge to any point in $X$.How do I show this,here I am stuck.Can someone please help me?
 A: Firstly, a minor adjustment: you can assume WLOG that the $A_n$ are not only closed, but increasing too: $A_n \subseteq A_{n+1}$ for all $n$. This is because finite unions of closed nowhere dense subsets are still closed nowhere dense and we can replace $A_n$ by $A'_n:=\bigcup_{i=1}^n A_i$ and we keep the same union and gain increasingness and so decreasingness of the $B_n$.
Secondly, make sure that the balls you're using are the closed balls of radii $< \frac{1}{n}$ at stage $n$ and ensure that each of closed balls lies inside the previous closed ball. Then the Cantor theorem on decreasing closed sets with diameters going to $0$ can be used and we have a point in the intersection (appealing to the completeness of $(X,d)$) and as each ball at stage $n$ avoids $A_n$, a point in the intersection cannot be in any $A_n$, which gives the contradiction we need. So the contradiction then comes from using completeness, not from refuting completeness by finding a wrong Cauchy sequence.
Cantor's theorem is the complete metric variant found here. 
