Let $S\subseteq\mathbb{R}$ be an arbitrary infinite and bounded set. We assume that $S$ has at least one limit point $p$, i.e. in every neignourhood of $p$ we find a $x\in S$ with $x\neq p$.
Show that this implies the least upper bound property, i.e. every bounded non-empty set has a supremum.
My approach:
Let's consider an arbitrary infinite non-empty bounded $T\subseteq\mathbb{R}$ and let be $p$ a limit point of $T$.
1.) If there exist finite many $x\in T$ such that $x>p$, then the greatest of those $x$ is trivially the supremum and we are done.
2.) If there exist infinite many $x\in T$ such that $x>p$, then I tried to define strictly monotone sequeuences to prove the existence of a specific limit point which might be the supremum but I failed so far. Maybe someone can give me a hint how to proceed?
You may have confused yourself with your notation. You didn’t introduce the variable $S$, but you seem to be using it to denote the bounded non-empty set whose supremum we want – that might lead to confusion because $S$ was already used in the theorem quoted at the top, and we want to use that theorem in the proof. So to keep it clear I’ll denote the bounded non-empty set whose supremum we want by $T$.
You can’t start by assuming that $T$ has a limit point, since it’s not given that $T$ is infinite, whereas $S$ in the theorem that ensures the existence of a limit point is assumed to be infinite.
Instead, you could proceed as follows. $T$ is non-empty, so it has an element $x$. $T$ is bounded, so there is an element $y$ greater than all $x\in T$. Construct an ascending and a descending sequence by halving the distance between $x$ and $y$ in each step and replacing the appropriate one of them by the midpoint. You should be able use the quoted theorem to show that both of these sequences have a limit point; and because the distance between them converges to zero, these two limit points must coincide. Then you should be able to show that this coinciding limit point is the supremum of $T$.