This question arises proving exercise 1.7.4 of Ryszard Engelking General Topology. In particular, traying to show that every linearly ordered sequential space is first-countable. The proof is given in Meyer's article Sequential properties of ordered topological spaces (avaible here for free). I can follow the reasing of the proof more or less. My problem comes later, in the proof of Lemma 1 that he uses to study only some particular $\mathfrak m$-nets ($\mathfrak m$ refers to the cardinality of the directed set of the net).
Lemma 1. Let $X$ be an ordered topological space and $A$ a subset of $X$. If an $\mathfrak m$-net in $A$ converges to $x\in X\setminus A$, then there is a strictly monotone $\mathfrak m$-net in $A$ which is directed by an ordinal and converges to $x$.
The proof goes as follows: Let $\{x_n\}$ be a sequence in the hypothesis and define
$$ A_1=\{x_n: x_n<x\} \qquad \mbox{ and } \qquad A_2=\{x_n:x<x_n\} .$$
Either $x$ belongs to the sequential closure of $A_1$ or to the sequential closure of $A_2$ (or both). We suppose the former case.
Now, since $A_1$ is a lineraly ordered set, it is directed, so the inclusion $\iota:A_1\rightarrow X$ is a net. Moreover, he says that this net converges to $x$. And here it is where my problem comes: I'm sure it does, it is intuitive, but I'm not sure how to prove it. That is what I've tried (Im' only inetrested in sequences although my attempt should work for an arbitrary net):
My attempt. I assumed that $\iota:A_1\rightarrow X$ did not converget to $X$, traying to find some contradiction.
Let me denote by $\{y_a\}$ the sequence $\iota:A_1\rightarrow X$. Now, if $\{y_a\}$ doesn't converge to $x$, there exists a neighbourhood $V$ of $x$ such that, for each $a\in A$ there exists $a'\in A$, $a<a'$, such that $y_{a'}\notin V$. But repeating this argument, I could find an infinite sequence $\{y_{a_n}\}$ of elements of $A_1$ that are not in $V$, which seems to be an absurd, because $\{y_a\}$ was just a rearrangement of $\{x_n\}$ and we know that only finitely many $x_n$ didn't belong to $V$.
Is my proof correct?
By the way, now Meyer invokes the Zorn lemma to find a well-ordered cofinal subset of $A_1$. The inclusion $A_3\hookrightarrow X$ is the $\mathfrak m$ he is looking for. But I'm interested inly in sequences. Do I need Zorn lemma too or there is another way? Thanks.
For sequences things are much easier in Lemma 1: every sequence in an ordered set has a strictly monotonous or constant subsequence (strictly increasing or strictly decreasing or constant).
Stated rigorously: If $(X,<)$ is linearly ordered and $(x_n)$ is a sequence in $X$ then there is an infinite subset $M \subseteq \mathbb{N}$ and $R \in \{<,>,=\}$ such that for all $n,m \in M: x_n R x_m$ holds.
Proof: immediate from the infinite Ramsey theorem using the colouring of pairs of integers given by $\{n,m\}$ has colour 1 when $x_{\min(n,m)} < x_{\max(n,m)}$, 2 if $x_n = x_m$ and 3 if $x_{\min(n,m)} > x_{\max(n,m)}$. Then the promised homogeneous subset defines (enumerated) one of the three types of subsequence, depending on the colour. (For a proof without the nice tool of Ramsey theory, see e.g. this, variations of which can be found in a lot of online places (and in books))
So when $x_n \to x$ converges and all $x_n \in A$ and $x$ is not, the constant sequence option is ruled out, and we immediately get a monotonous subsequence with the same limit.