Let $X$ be a first-countable space, and let $\mathcal B_x=\{B_1,B_2,\dots\}$ be a local base at $x\in X$ satisfying $B_1\supset B_2\supset\dots$. Given a net $(x_\alpha)_{\alpha\in A}$ that converges to $x$, I want to show that it has a convergent subsequence (that is, a subnet whose index set is $\mathbf N$). In more detail, we want to find a monotone cofinal map $\phi\colon\mathbf N\to A$ such that $(x_{\phi(n)})_{n\in\mathbf N}$ is a sequence that converges to $x$. I tried to use the convergence of the net to define a sequence as follows. Choose $\phi(1)\in A$ such that $x_\alpha\in B_1$ whenever $\alpha\ge\phi(1)$. Then choose $\phi(2)\in A$ such that $\phi(2)\ge\phi(1)$ and $x_\alpha\in B_2$ whenever $\alpha\ge\phi(2)$. Continuing in this way, we define a monotone sequence $\phi$. But I am stuck here — I do not know how to prove that $\phi$ is cofinal (and I am not even sure if it is).
How should I proceed? Thank you.
This is not true. For instance, you could just take a constant net with value $x$ with an index set that has no countable cofinal subset (e.g., $\omega_1$). Such a net has no subsequences at all.
There are less trivial examples as well, which show there is no easy fix by disallowing (say) nets that are eventually in the intersection of all neighborhoods of $x$. For instance, the index set of our net could be $\omega\times\omega_1$ with the product order, with $x_{(n,\alpha)}\in B_n\setminus B_{n+1}$ for all $(n,\alpha)\in\omega\times\omega_1$. Such a net will converge to $x$, since it is eventually in each $B_n$. However, this net still has no subsequences at all: no countable subset of the index set is cofinal, since any countable subset is bounded on the second coordinate.