Product of sets and supremum
Let $\alpha = \sup A$ and $\beta = \sup B$.
For every $a\in A$ and $b \in B$ we have
$$ab \leq \sup_{b\in B} a b = a \beta \leq \sup_{a\in A} a \beta = \alpha \beta,$$
so we have $\sup AB \leq \alpha\beta$.
Now let $(a_n)_{n\in \mathbb N} \subset A$ and $(b_n)_{n \in \mathbb N} \subset B$ be sequences such that $a_n \to \alpha$ and $b_n \to \beta$ as $n \to \infty$.
It is then clear, that $(a_nb_n)_{n \in \mathbb N} \subset AB$ and $a_nb_n \to \alpha\beta$ as $n \to \infty$, so $\sup AB \geq \alpha\beta$ and therefor we have $\sup AB = \alpha \beta$.