Ok...
(1) Suppose $\beta < \alpha$, since $\alpha = \sup A$, any number $t < \alpha$ is not an upper bound of $A$ by the definition of the least upper bound
(2) But $\beta$ is an upper bound of $B$, so $\forall x \in B$, $x \leq \beta$, in particular, every $x\in A, x \leq \beta$ so that...