Show that $\sup\ A < \sup\ B \implies \exists b\in B \ |\ b \ \text{is upper bound for} A$
Let $\gamma = \sup\ B - \sup\ A > 0$ and $\epsilon = \frac{\gamma}{2}> 0$ further set $b=\sup\ B- \epsilon$, this should force $b\in B$. At the same time we have $b=\sup\ B- \epsilon=\sup\ A + \gamma-\epsilon=\sup\ A + \frac{\gamma}{2}$. So $b$ is an upper bound of $A$.
Is this line of reasoning valid or am I missing something?
P.S this is exercise 1.3.9 in Understanding Analysis by Abbott.
You can't assume $\sup B-\epsilon$ is in $B$ as Brian pointed out. Consider for example $B=\{3\}$ and $A=\{2\}$. Then $\gamma=1$ and $\sup B-\epsilon=2.5\not\in B$.
Instead we can argue that $\sup A$ is strictly less than the least upper bound of $B$, so by definition it cannot be an upper bound of $B$, so there is some $b\in B$ with $b>\sup A\geq a$ for all $a\in A$, thus $b$ is an upper bound for $A$.