Let be $\Omega$ an uncountable set and $\leq$ a linearly ordered relation on $\Omega$. Than can we say that exists a co-initial sequence and cofinal sequence in $\Omega$?
Where we say:
a subset $B$ of $\Omega$ is said to be coinitial if it satisfies the following condition: For every $a \in \Omega$, there exists some $b \in B$ such that $b \le a$.
A subset $B$ of $\Omega$ is said to be cofinal if it satisfies the following condition: For every $a ∈ \Omega$, there exists some $b ∈ B$ such that $a \leq b$.
This is easiest using transfinite recursion.
Fix a well-ordering of $\Omega$, now proceed by picking a random element, then at each step if you're still bounded, pick a larger/smaller element. Until you've exhausted all the elements or got an unbounded sequence.
You can also use Zorn's lemma when you order well-ordered (or anti-well ordered) subsets by end-extension ($A\unlhd B$ if $B\cap(-\infty,\sup A)=A$). It is not hard to show the conditions for Zorn's lemma hold, and therefore there is an unbounded subset of $\Omega$ which is well-ordered under $\leq$.
The axiom of choice is necessary, by the way, as it is consistent that there is a subset of $\Bbb R$ which is unbounded, but has no countably infinite subset. Looking at the induced linear order from $\Bbb R$ you get an uncountable set, with a linear order, but every sequence is finite and therefore bounded.