Given any infinite totally ordered set $T$, does there exist a countable subset $C$ such that $C$ is not upper bounded in $T$?
Edit: $T$ also is unbounded from above. My bad for not including this first.
This statement is non-trivial for large $T$, and I don't really have a grasp on how to prove it (my inclination says it is probably true since every simple example that comes to mind satisfies the condition).
Any pointers would be much appreciated. I'm sure this is far simpler than I'm making it out to be.
It depends on the cofinality of $T$, i.e., the smallest cardinality of a cofinal subset of $T$. If $\operatorname{cf}T>\omega$, then every countable subset of $T$ is bounded above; $\omega_1$, the first uncountable ordinal, is the simplest example. If $\operatorname{cf}T=\omega$, let $\langle x_n:n\in\omega\rangle$ be a cofinal sequence; clearly $\{x_n:n\in\omega\}$ is a countable subset of $T$ with no upper bound. The only other possibility is that $T$ has a maximum element, in which of course every non-empty subset of $T$ is bounded above, but that possibility is excluded by hypothesis.