Let $(S,\leq)$ be a totally ordered set. We say that $E\subseteq S$ is left-finite if, for every $x\in S$, the set $E\cap\{y\in S : y\leq x\}$ is finite. Every finite subset is for example left-finite. A less trivial example is the set of natural numbers inside the real ones: each real number has only finitely many natural numbers before it. Does every totally ordered set without a greatest element have an infinite left-finite subset?
It is easy to see that every left-finite subset $E$ of a totally ordered set $(S,\leq)$ is at most countable. The function $f:E\rightarrow\mathbb{N}$ given by \begin{equation} f(e)=\#\bigl(E\cap\{y\in S: y\leq e\}\bigr) \end{equation} is indeed strictly increasing, hence one-to-one. This also shows that $E$ has a smallest element. The question can be therefore rephrased as follows: does every totally ordered set without a greatest element contain an increasing sequence without an upper bound?
No: consider the first uncountable ordinal $\omega_1$. Any countably infinite subset $E$ of $\omega_1$ has an upper bound in $\omega_1$, so $E$ is not left-finite.