Proof of countable intersection of dense and open sets in $\mathbb{R}$ I wanted to see if my proof of the Baire Category Theorem in $\mathbb{R}$ is correct.
Let $\lbrace D_n \rbrace$ be a collection of dense, open subsets of $\mathbb{R}$.
We want to show every open $W \subset \mathbb{R}$ intersects $\bigcap_{1}^{\infty}D_n$.
To that end, let $W$ be non-empty, open, and take $x_1$ and $\epsilon_1 > 0$ in $W \cap D_1$ s.t $B_{\epsilon_1}(x_1) \subset W \cap D_1$. We can do this because $W \cap D_1$ is also non-empty and open. This ball is open, and hence its intersection with $D_2$ is non-empty, open. So, we can choose $x_2$ and $\epsilon_2 < \frac{{\epsilon}_{1}}{2}$ such that $B_{\epsilon_2}(x_2) \subset B_{\epsilon_1}(x_1) \cap D_2$.
Similarly, for $n \in \mathbb{N}$, choose $x_n$ and ${\epsilon}_n < \frac{{\epsilon}_{n-1}}{2}$ s.t $B_{\epsilon_n}(x_n) \subset B_{\epsilon_{n_1}}(x_{n-1}) \cap D_n$.
Then, $\lbrace x_n \rbrace$ is Cauchy, hence convergent, say to $x$. Then, $x$ will be in $\bigcap_{1}^{\infty}D_n \cap W$.
 A: (I don't know if I should let this be an answer or a comment since it's a "long comment" so I'll just make this an answer)
The proof seems mostly valid as-is (and this version can be generalised to complete normed $\mathbb{R}$-vector spaces), however the "interesting" part of the proof (at least to me) is showing that $(x_n)_n$ is a Cauchy/convergent sequence, so detailing that part at least a little might be worth it. You chose to divide by $2$ the $\varepsilon_n$ after each step so I suspect you already know why and how it works but still.
Moreover, to make sure the limit $x$ is in every $D_n$, you might want to enforce/emphasize having the closed ball - let's denote it with $B'_{\varepsilon_n}(x_n)$, I don't know the usual English notation for a closed ball - be included in $B_{\varepsilon_{n-1}}(x_{n-1}) \cap D_n$ during the construction and not just the open ball.
Much like for showing $(x_n)_n$ was convergent, detailing that part of the argument should tell you why the closed ball is pretty much needed here.
I don't have anything to say for the rest though, it's clear, concise and shows what needed to be showed!
EDIT: Wanted to go back on this. I did overcomplicate things, as I thought, by requiring summation when it's not necessary. I feel very silly right now...
Your proof can be generalized to all complete metric spaces by using the fact that the decreasing intersection of countably many closed subsets such that the sequence of their diameters converges to $0$ is of the form $\{y\}$ (and there is a version of it in $\mathbb{R}$ that you could use without all the machinery of complete metric spaces).
So the choice of the speed of convergence towards $0$ of $\varepsilon_n$ wouldn't matter with that argument, however it still does without that argument, and what I said about closed vs open balls is still relevant in both cases.
