Why does there exist a such maximal chain in order topology? Hocking and Young-Topology p.54
Let $X$ be a topological space equipped with an order topology and $x\in X$.
Define $S=\{(a,b): x\in (a,b)\}$ and assume that $S$ is nonempty.
How do I show that there exists a maximal chain $\mathscr{C}$ in $(S,\subset)$ such that $\bigcap \mathscr{C}=\{x\}$?

This isn't true.  For instance, let $X=\omega\sqcup \{x\}\sqcup\omega_1^*$, where $\omega_1^*$ denotes $\omega_1$ with the reverse of the normal order, ordered such that for all $n\in\omega$ and $\alpha\in\omega_1^*$, $n<x<\alpha$.  Suppose you have any chain $C$ of open intervals such that $\bigcap C=\{x\}$.  Then for each $n\in\omega$, we can choose an interval $(m_n,\alpha_n)\in C$ for some $m_n\in\omega$ which is larger than $n$ and some $\alpha_n\in\omega_1^*$.  If $(a,b)\in C$ were contained in every $(m_n,\alpha_n)$, then $a$ would have to be greater than every $m_n$, so $a\geq x$, so $x\not\in (a,b)$.  Thus no such $(a,b)$ can exist.  Since $C$ is a chain, this means every element of $C$ contains $(m_n,\alpha_n)$ for some $n$, so $\bigcap C=\bigcap (m_n,\alpha_n)$.  But there exists $\beta\in \omega_1^*$ such that $\beta<\alpha_n$ for all $n$, and then $\beta\in \bigcap (m_n,\alpha_n)$, which is a contradiction.
More generally, this statement will be true if and only if $x$ has an immediate successor, $x$ has an immediate predecessor, or the cofinality of $x$ from the left and the right are equal.
