Supremum of non-empty subset of $\mathbb{R}$

Question

Let A be a non-empty subset of $\mathbb{R}$ that is bounded above and put $s=\sup A$
Show that if $s\notin A$ the the set $A\cap (s-ε,s)$ is infinite for any $ε>0$

This has to be solved using contradiction, by supposing $A\cap (s-ε,s)$ is an finite set. But I am not sure how to proceed after this.

Answer 1

Suppose $(s-\epsilon,s)\cap A$ is not infinite, then the set $(s-\epsilon,s)$ has a maximum, call it $M$. Notice $M$ is an upper bound for $A$. To see this suppose $M$ is not an upper bound, then there is $a\in A$ with $a>M$, but then $a<S$ and $a>M>S-\epsilon$, so $a$ is in $(s-\epsilon,s)\cap A$. This contradicts that $M$ is a maximum for $(s-\epsilon,s)\cap A$. We have established $M$ is an upper bound for $A$, but this is a contradiction, because $M<S$ and $S$ is supposedly the least upper bound for $A$.

Answer 2

HINT

Suppose that $A\cap (s - \varepsilon_0, s)$ is finite for some $\varepsilon_0$ and consider two cases: it can be empty or it is not, and in the latter case, consider its maximum element (every finite subset of an ordered set has a maximum!). Both cases lead to contradictions.