Two variants of proof for Baire's category theorem

One of the variants of the so called Baire's Category Theorem (BCT) says that

Given a (possibly nonempty) complete metric space $(X,d)$ and a system of open dense subsets $(O_n)$ of $X,$ then $\bigcap_{n \in \mathbb{N}} O_n$ is dense in $X.$

Most proofs of this theorem goes like this: to show that the intersection of $O_n$'s is dense, it is enough to show that for any $y \in X,$ $\bigcap_{n \in \mathbb{N}} O_n \cap B_{r}(y)$ with $B_{r}(y)$ an open ball of radius $r$ and centered at $y,$ is nonempty. Since $O_1$ is dense in $X,$ the intersection $O_1 \cap B_r(y)$ is nonempty and we can choose $x_1$ and $r_1 < r/2$ such that $B_{r_1}(x_1) \subset O_1 \cap B_r(y).$ In fact, by shrinking $r_1,$ we can arrange things in such a way that $\overline{B_{r_1}(x_1)} \subset O_1 \cap B_r(y).$ Then, since $O_2$ is dense, we can find $x_2 \in B_{r_1}(x_1) \cap O_2$ such that $\overline{B_{r_2}(x_2)} \subset O_2 \cap B_{r_1}(x_1)$ with $r_2 < r/4.$ Going this way, we recursively define sequences $(x_n)$ and $(r_n)$ such that $\overline{B_{r_{n+1}}(x_{n+1})} \subset O_{n+1} \cap B_{r_n}(x_n)$ with $r_{n+1} < r/2^{n+1},$ showing that $(x_n)$ is Cauchy with a limit $x.$ Since $\overline{B_{r_n}(x_n)}$ is closed and $x_i$ lies in $\overline{B_{r_n}(x_n)}$ for every $i>n$ with $n \in \mathbb{N}$ arbitrary, it follows that $x$ lies in $\overline{B_{r_n}(x_n)}$ for every $n,$ hence it lies in every $O_n$ and it lies also in $B_r(y),$ which completes the proof.

A slightly more straightforwardly looking variant of this proof is given for example here, the only difference is that it gives the sequences $(x_n)$ and $(r_n)$ in such a way that we have only $B_{r_{n+1}}(x_{n+1}) \subset O_{n+1} \cap B_{r_n}(x_n)$ (without ever taking the closure of the balls). The author then goes to the conclusion that, due to its construction, the limiting point $x$ (which no doubt exists, since $X$ is complete), lies in $\bigcap_{n \in \mathbb{N}} O_n \cap B_r(y).$ Is this conclusion correct? If so, why?

My understanding is that essentially the same argument with the closures must be somehow implicitly involved, or is there another way that bypasses the use of closures (like for example contradiction with cauchyness of the sequence involved)?

Let's consider the example $$X=[0,1], d(x,y)=|x-y|, O_n = (0,1).$$
Then $O_n$ is open and dense for every $n\in\mathbb N$.
However, we can assume that the construction of $x_n$ and $r_n$ yields $$x_n= 2^{-n}, r_n = 2^{-n}.$$
Then $B_{r_{n+1}}(x_{n+1}) \subset O_n \cap B_{r_n}(x_n)$ is true, (because $B_{r_{n+1}}(x_{n+1})\subset (0,1)$) as well as $r_n<\frac1n$ as required in the proof that you linked. However, the limit is $x=\lim x_n = 0$, which is not in the intersection of $\cap_{n\in\mathbb N} O_n = (0,1)$.