About the theorem for the first-countable space I'm learning basic topology in my university. And when learning about the first-countable space, my teacher told us that for every first-countable space, there's a very important theorem like this.


If $X$ is a first-countable space, then with every $x \in X$, there exists a "regular neighborhood base" $(U_n)$, meaning that:
     1) Every $U_n$ is an open set contains $x$.
     2) If $x_n \in U_n$ for each $n$, then the sequence $(x_n)$ converges to $x$.


But even after reading some book about topology(like of Munkres, Folland and searching on the Internet), I can't find any theorem like that. So I wonder if there really exists a theorem like this, or there's some mistake in the theorem. Is first-countable condition enough for the theorem? If anyone knows about this theorem, please cite some resource for me to understand more. Thanks so much.
 A: If $X$ is a first countable topological space, and $x_0\in X$, there exists a countable collection $\{G_n\}_{n=1}^{\infty}$ of open sets $G_n$ with the property that every open set $U$ containing $x$, contains $G_n$ for some $n$. Such a collection is also called a local base at $x_0\in X$.
Now once you have a local base at $x_0\in X$, you can construct a nested local base at $x\in X$. That is, you can find a local base $\{B_n\}_{n=1}^{\infty}$ of open sets at $x\in X$ with the additional property that it is nested, i.e. $B_{n+1}\subset B_n$ for all $n$. To see this, just take any local base $\{G_n\}_{n=1}^{\infty}$ at $x_0$ and put
$$B_1=G_1,\enspace B_2=G_1\cap G_2,\dots B_n=\bigcap_{k=1}^n G_k$$
Each $B_n$ is open as an intersection of finitely many open sets, and clearly each $B_n$ contains $x_0$, so $\{B_n\}_{n=1}^{\infty}$ is a nested local base at $x_0$.
Now if you choose a point in each one of the members of a nested local base at $x_0$, say $x_n\in B_n$, the sequence $x_n$ must converge to $x_0$. This is because if $G$ is any open set containing $x_0$, then by definition of a local base there exists some $N$ such that $B_N\subset G$, and since our local base is also nested, i.e., $B_n\subset B_N$ for all $n>N$, we will have that $x_n\in G$ for all $n>N$, and so $\lim_{n\to\infty}x_n=x_0$.
Finally, you will not find this fact stated as a theorem in every book on topology, because many of them will give it as an exercise, or a remark.
