For this question, let us assume $S$ is an uncountable nonempty ordered set such that every its nonempty subset has an infinum and a supremum. And let us assume that $f \colon S \to S$ is a monotonically increasing function. And suppose that $f(x) \neq x$ for all $x \in S$. Now to the question matter.
We have $inf(S) < f(inf(S))$. Define $a_{0,0} = inf(S)$, $a_{0,n+1} = f(a_{0,n})$. We must have $sup\{a_{0,n}\} < f(sup\{a_{0,n}\})$. Proceed inductively, by defining $a_{1,0} = sup\{a_{0,n}\}$, $a_{1,n+1} = f(a_{1,n})$. Now define $\Omega_n = a_{n,0}$.
Question: I guess, we cannot infer that it is not the case that $sup\{\Omega_n\} < f(sup\{\Omega_n\})$ from how $\{\Omega_n\}$ was defined (or anything whatsover), since the set $S$ is uncountable and induction is a bad weapon to tackle this, am I right? To clarify: the sequences $\{a_{m,n}\}$ are constructed in a such manner that every consecutive supremum, obtained with $f$, is tackled by the next sequence $\{a_{m+1,n}\}$; so I wondered, if the definition of $\{\Omega_n\}$ is enough to get rid of, as I called it, "the next supremum"?
The original problem, found in Berkeley Supplements for Rudin's Principles of Mathematical Analysis on page 4, is stated as follows. The problem is about fixed points, you can easily find it on this site answered (once, I saw it right to this question in the right column of the related questions); but I won't search it myself, since then I will inadvertently see the solution, which will spoil my wish to solve it myself.
Let $S$ be a nonempty ordered set such that every nonempty subset $E \subseteq S$ has both a least upper bound and a greatest lower bound. (A closed interval $[a, b]$ in $\mathbb R$ is an example of such an $S$.) Suppose $f \colon S \to S$ is a monotonically increasing function; i.e., has the property that for all $x$, $y \in S$, $x \leq y \implies f(x) \leq f(y)$. Show that there exists an $x \in S$ such that $f(x) = x$.