I have to prove
$X$ connected in the order topology (w.r.t. a linear order <) $\Rightarrow$ every $A\subset X$ that has an upper bound has a supremum
My attempt:
Reason by contradiction: take $A\subset X$ with an upper bound but no supremum. It follows that $A$ has no maximal element, and that there are infinite increasing sequences $(x_1,x_2,...)$ in $A$, and an infinite sequence $(y_1,y_2,...)$ of upper bounds on $A$. We then consider $$A_<=\bigcup_{a\in A}\{x\in X|x<a\}$$ $$A_>=\bigcup_{a\text{ an upper bound on $A$}}\{x\in X|a<x\}$$ These are both open in the order topology. It is not hard to verify that these sets partition $X$ and that they are also closed. This way we exhibit $X$ as the union of $2$ separated components, which is a contradiction.$\square$
Since I'm self studying and have no answers of the exercises I just want to verify this answer. I'm most interested in the validity of my reasoning related to the upper-bound stuff (the existence of the infinite sequences and such). Thanks.
The sequences aren’t doing anything for you: you never actually use them. It’s the sets $A_<$ and $A_>$ that you want, and with them your argument is fine, though I’d say that it’s a little easier to observe that they are open and complementary.
The sequences always exist, but the increasing sequence may not be cofinal in $A$. For example, let $X$ be the closed long ray, and let $A$ be the open long ray; then $A$ is unbounded in $X$, but any strictly increasing ordinary sequence in $A$ has a supremum in $A$. Similarly, it’s possible that no decreasing sequence is coinitial in the set of upper bounds (e.g., just turn this example around).