Prove that the intersection have infinite elements

I want to prove this theorem, and I made a proof but I want to know your opinion...

Theorem: Let $A \subset \mathbb{R}$, where $A\neq \emptyset$, and $S=sup (A)$. Prove that if $S\notin A$, then $\forall \delta>0$ the set $A \cap (S-\delta,S)$ have infinite elements.

My proof:

First I'm going to prove that the intersection $A \cap (S-\delta,S) \neq \emptyset$, let's proceed by contradiction... Suppose that $A \cap (S-\delta,S) = \emptyset$ then there exists an $\alpha \in (S-\delta,S)$ s.t $\alpha \notin A$, then it satisfies that $a<\alpha<S, \forall a \in A$, this is a CONTRADICTION, because $S$ is the smallest upper bound (by the fact is the sup) and I found a smaller upper bound... Then the intersection is non empty...

Now I'm going to prove that $A \cap (S-\delta,S)$ have infinite elements proceeding by contradiction... Let's suppose that the intersection have a finite amount of elements, let's say $n$, all this elements are going to be labeled with $a_i$ with $i=1,...,n$ ordered from the smallest to the biggest one like: $a_1<a_2<...<a_n$, then it's clear that $a \leq a_n<S$ for all $a\in A$, this means that $a_n$ is the smallest upper bound of A, then $a_n$ is the $sup(A)$, and $a_n \in A$, this is a CONTRADICTION to the fact that $S$ is the $sup(A)$ and that $S \notin A$, then we get what we were looking for.

This is my proof... What's your opinion?

You don't need contradiction. Suppose that $A$ is novoid and $\sup(A)\not \in A$. Let $x_0 < \sup(S)$. Then there is some $x_1\in A$ with $x < x_1 < \sup(A).$. Assume we have chose $x_1 < x_2 \cdots x_n < \sup(A)$. Since $x_n < \sup(S)$ there is an $x_{n+1}\in S$ so that $x_{n+1} < \sup(A)$. By induction we have an increasing sequence $\{x_n\}$ with $x_n\in S$, $n\ge 1$.