Consider the Sequence Lemma:
Let $X$ be a topological space, $A\subseteq X$ any subset and $x\in X$. If there is a sequence of points in $A$ converging to $x$, then $x\in\bar A$; the converse holds if $X$ is first-countable.
In the proof of the converse provided here they define a sequence of the elements of the neighborhood basis $\mathcal{U}$ of $x$ as $\{U_i\}_{i \in \mathbb{N}}$. Then they define the sequence of non-empty sets $\{S_i\}_{i \in \mathbb{N}}$ as $S_i = A \cap (\cap_{k=1}^i U_k)$. Finally they create the sequence $\{x_i\}_{i \in \mathbb{N}}$ where $x_i \in S_i$, and as $x_i \rightarrow x$ then the proof is complete.
As far as I understand the first-countability condition is used to assert the existence of a bijective function $\sigma: \mathbb{N} \rightarrow \mathcal{U}$ which we can use to create the sequence $\{U_i\}_{i \in \mathbb{N}}$ by making $U_i = \sigma(i)$. I do not understand why the first-countability condition is needed, as I will explain now. Consider the case where $\mathcal{U}$ is uncountable, and let $\mathbb{U}$ be an uncountable well-ordered set. Then we take a bijective function $\omega: \mathbb{U} \rightarrow \mathcal{U}$ and use it to define the sequence $\{U_i\}_{i \in \mathbb{U}}$ as $U_i = \omega(i)$. The rest of the proof follows in the same way.
My question is: What is the flaw in my reasoning, and why?
Thanks in advance!
Usually, in the context of topology, a "sequence" is defined to be indexed by $\mathbb{N}$, not by an arbitrary well-ordered set. So if your $\mathbb{U}$ is different from $\mathbb{N}$, what you are constructing is simply not a sequence at all.
Even if your definition allows sequences indexed by any well-ordered set, though, the argument still does not work. The issue is that you know that $S_i = A \cap (\bigcap_{k=1}^i U_k)$ is nonempty because $\bigcap_{k=1}^i U_k$ is a neighborhood of $x$ and $x\in\overline{A}$. However, if $i$ is an element of your uncountable well-ordered set $\mathbb{U}$, there may be infinitely many $k\leq i$, so you would be taking an infinite intersection of neighborhoods of $x$ which may no longer be a neighborhood of $x$. Consequently, $S_i$ could be empty and there could be no way to pick $x_i$.