$X$ is a well ordered set. $S_\Omega= \{ x:x\in X\text{ and } x<\Omega\}$ such that it is the minimal uncountable well ordered set. The section $S_\Omega$ of $X$ is uncountable and any other section of $X$ is countable (a section is $S_\alpha=\{ x:x\in X\text{ and } x<\alpha\}$).
I'm having trouble with open sets with the order topology. I've read that if $x$ has no predecessor then $\{x\}$ is not open, and if $x$ has a predecessor then $\{x\}$ is open. Why does this happen?
If $x$ has predecessor $y$ then $\{x\}=\{z\mid z<x+1\}\cap\{z\mid y<z\}$ or more shortly $\{x\}=(y,x+1)$ which is an open set according to the order topology.
If $x$ has no predecessor then for every $y<x$ the set $(y,x+1)$ contains $y+1<x$ so that $\{x\}\neq(y,x+1)$.
Edit:
One exception though: if $x$ is smallest element of $X$ then it has no predecessor but $\{x\}$ is open.