An "if and only if" proof regarding upper bounds

Question

Is my proof okay? If not, what should I change?

Answer 1

I will write down one of the implications properly:

You want to show that the first statement (Let me call it $P1$)

$$\forall_{u\in \mathbb{R}} \exists_{x\in A} : x>u$$

Is equivalent to the second statement (P2)

$$\forall_{\varepsilon>0} \exists_{x\in A} : x>S+\varepsilon$$

This is how you prove that $P1\Rightarrow P2$:

Let $\varepsilon>0$ be any real number. Choose $u=S+\varepsilon$ then by (P1) we know that there exists $x\in A$ such that $x>u$, hence $x>S+\varepsilon$.

Thus, we showed that for every $\varepsilon>0$ there exists $x\in A$ such that $x>S+\varepsilon$ (which is exactly (P2)).

I leave the case $P2\Rightarrow P1$ to you.