In the proof of The Baire Category Theorem for a complete metric space $ (X,d)$, it is shown that the intersection of a countable collection $ \{ A_k \}_{k=1}^{\infty} $ of open dense sets is dense. The proof as given on the wikipedia page: Wiki Proof starts with an arbitrary point $ p \in X$ and a basic open neighborhoood $ B(p,\epsilon)$ of $p$ and shows that an infinite sequence of closed nested balls $ \{ \overline{B}(x_k,\epsilon_k )\}_{k=1}^{\infty} $ can be constructed, each of which is contained in $ B(p,\epsilon)$ and where for each $n$, $ \overline{B}(x_n,\epsilon_n) \subseteq (\cap_{k=1}^{n} A_k) $ and $\lim_{k \to \infty}\epsilon_k \to 0$.
The part that bothers me is where one shows that given an arbirary finite sequence of such closed nested balls, you can lengthen it by one. By induction you have arbitrarily long finite chains of such closed nested balls but I understand you need choice to get an infinite one.
I would like to at least understand how to make this step explicit, as it seems to bear resemblance to other steps in proofs in analysis (Urysohn's lemma and proof that a compact metric space is sequentially compact). However, my attempt is really clumsy and possibly incorrect. The proof is shown below:
Let $ \{ A_{n} \}_{n=1}^{\infty} $ be a countable collection of open dense sets in a complete metric space $ (X,d) $. Let $ p \in X $ and let $ B(p,\epsilon) $ be a basic open nbhd of $ p $. We define $ (\mathcal{A}, \leq) $ to be a partially ordered set where $ \mathcal{A} = \{ (x,\overline{B}(x,\epsilon_{x})): \overline{B}(x,\epsilon_{x}) \subseteq B(p,\epsilon), \epsilon_{x} \in \{1/n : n \in \mathbb{N} \} \} $. For each $ (x,\overline{B}(x,\epsilon_{x})) $, we define $ \mu \colon \mathcal{A} \to \mathbb{N}_{0} $, by $ \mu((x,\overline{B}(x,\epsilon_{x}))) = \max \{ n \colon \forall k \leq n, \overline{B}(x,\epsilon_{x}) \subseteq A_{k} \} $.
Then define: $ (u,\overline{B}(u,\epsilon_{u})) < (v,\overline{B}(v, \epsilon_{v})) $, if $ \overline{B}(v,\epsilon_{v}) \subset \overline{B}(u,\epsilon_{u}) $, $ \mu( (v,\overline{B}(v,\epsilon_{v})))) > \mu( (u,\overline{B}(v,\epsilon_{u})))) $ and $ \epsilon_{v} < \epsilon_{u} $. Then this defines a partial ordering on $ \mathcal{A} $. Also $ \mathcal{A} $ is not empty, since $ A_{1} $ is open and dense and a metric space is regular, $ \exists x \in A_{1} \cap B(p,\epsilon) $ and there exists $ n \in \mathbb{N} $ such that $ x \in \overline{B}(x,1/n) \subseteq A_{1} \cap B(p, \epsilon) $. Now suppose that every chain in $ \mathcal{A} $ is eventually constant. Then each chain is bounded by its largest element and so by Zorn's Lemma there is a maximal element $ (x_{n}, \overline{B}(x_{n},\epsilon_{n}) ) $. If $ \overline{B}(x_{n},\epsilon_{n}) \subseteq \bigcap_{i=1}^{\infty} A_{i} $ then we are done. Otherwise, Let $ m = \max \{ k : \forall j \leq k, \overline{B}(x_{n},\epsilon_{n} ) \subseteq A_{j} \} $. Then $ \overline{B}(x_{n},\epsilon_{n}) \not \subseteq A_{m+1} $, but since $ A_{m+1} $ is open and dense and $ (X,d) $ regular, we can find some $ x_{n+1} $ and $ \epsilon_{n+1} $ such that $ \epsilon_{n+1} < \epsilon_{n} $ and $ x_{n+1} \in \overline{B}(x_{n+1},\epsilon_{n+1}) \subseteq \overline{B}(x_{n},\epsilon_{n}) \cap A_{m+1} $. This contradicts the claim that $ (x_{n},\overline{B}(x_{n},\epsilon_{n})) $ was maximal and so there must be an infinite chain which contains a stricly increasing infinite subchain $ \{ (x_{n},\overline{B}(x_{n},\epsilon_{n})) \}_{n=1}^{\infty} $. WLOG we take $ \overline{B}(x_{n},\epsilon_{n}) \subseteq A_{k} $ for all $ k \leq n $ and $ \epsilon_{n} \leq 1/n $.
Since the balls are nested and $ \epsilon_{n+1} < 1/(\epsilon_{n}+1) $, it is straitforward to prove that $ \{ x_{n} \}_{n=1}^{\infty} $ is Cauchy. Since $ X $ is complete $ x_{n} \to v $ for some $ v \in X $. Let $ n \in \mathbb{N} $. Then for all $ m \geq n $, $ x_{m} \in \overline{B}(x_{n},\epsilon_{n}) \subseteq A_{n}$. since $ \overline{B}(x_{n},\epsilon_{n}) $ is closed $ \lim_{k \to \infty}x_{n} = v \in \overline{B}(x_{n},\epsilon_{n}) \subseteq A_{n} $. Therefore, $ v \in B(p,\epsilon) \cap (\bigcap_{k=1}^{\infty} A_{k}) $ and so $ \bigcap_{k=1}^{\infty}A_{k} $ is dense.