Solutions

a) For contradiction, suppose that $\sup A>y$ for some $y\in B$. This means that there exists $x\in A$ such that

$$y<x<\sup A,$$

which clearly contradicts the premise that $x\le y$ for all $x\in A$ and all $y\in B$.

b) Since $\sup A\le y$ for all $y\in B$, $\sup A$ is a lower bound of $B$. Hence, by definition, $\sup A\le\inf B$.