Firstly, I realise that my question has already been answered a few times, however I just wanted to verify the proof that I came up with for Exercise 1.1.6 in Lebl's Basic Analysis I:
Let $S$ be an ordered set and $A \subset S$ be non-empty and bounded above. Suppose that $\sup A \notin A$. Show that $A$ contains a countably infinite subset.
What I have so far for a proof is:
Consider some $a_1 \in A$, then we know by the definition of the supremum, that $a_1 < \sup A$. Now pick some arbitrary $a_2 \in A$ where $a_1 < a_2$. We then know again by the definition of supremum that $a_1 < a_2 < \sup A$. Inductively, assume that $a_1, a_2, ..., a_n \in A$ with $a_1 < a_2 \cdots < a_n < \sup A$. By the definition of the supremum, we then pick an $a_{n+1} \in A$ such that $a_n < a_{n+1}$, meaning it follows that $a_{n+1} < \sup A$ by the supremum definition. Thus $a_1 < a_2 < \cdots < a_n < a_{n+1} < \sup A$. Therefore we can define an infinite set $B = \{a_1, a_2, ..., a_n, ...\}$, clearly since we were able to assign a natural number to each element of $B$, we know that $|B| = |\mathbb{N}|$, in other words $B \subset A$ is countably infinite.
Furthermore, is it valid to use this result to prove the result in Exercise 1.1.12 which asks for me to prove that every ordered field has a countably infinite subset? My logic is if I assume $F$ is an ordered field and let $A = \{ x \in F \ |\ x < a \}$, then I can use the result from 1.1.6 to show that $A$ has a countably infinite subset which is thus also a subset of $F$.