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Another way to write the proof is to use an alternative equivalent definition of supremum:


 > For a bounded set of real numbers $A$, $L=\sup A$ if 
>- $L$ is an upper bound for $A$, i.e., for all $a\in A$: $a\le L$;
>- any number smaller than $L$ is *not* an upper bound of $A$: for every $\epsilon>0$, there exists $a\in A$ such that $a>L-\epsilon$. 

As in the other answer, we let $L=(\sup A)(\sup B)$ and we show that

- $L$ is an upper bound for $A\cdot B$;

- for every $\epsilon>0$, there exists $a\in A$ and $b\in B$ such that
$$
ab> L-\epsilon\tag{1}
$$

The first statement is proved in the same way as in the other answer. Let us show the second one. Let $\epsilon >0$. We want to find $a$ and $b$ so that (1) holds. 

By the definition of supremum, for every $\epsilon'>0$, we can find $a\in A$ and $b\in B$ so that

$$
a+\epsilon'>\sup A,\quad b+\epsilon' >\sup B
$$
which gives $$ab+(a+b)\epsilon'+(\epsilon')^2>L\tag{2}$$

So, if we pick $\epsilon'$ so that
$$
\epsilon \ge (\sup(A)+\sup(B)+1)\epsilon'
$$
and correspondingly, the $a$ and $b$ that (2) holds, we have
$$
ab+\epsilon \ge ab +(\sup(A)+\sup(B)+1)\epsilon'\ge ab+(a+b+\epsilon')\epsilon'>L.
$$
We are done.