How can I show that $\sup(AB)\geq\sup A\sup B$ for $A,B\subset\mathbb{R}$ where $A\cup B$ is positive and bounded?

There's a simple way to prove the result. For all $0 < a\in A$, $0 < b\in B$ we have: $$\sup(AB)\geq ab\iff\frac{1}{a}\sup(AB)\geq b$$ hence $B$ is bounded above by $\frac{1}{a}\sup(AB)$ so $$\frac{1}{a}\sup(AB)\geq \sup B\iff \frac{1}{\sup B}\sup(AB)\geq a$$ hence $A$ is bounded above by $\frac{1}{\sup B}\sup(AB)$ so $$\frac{1}{\sup B}\sup(AB)\geq \sup A\iff \sup(AB)\geq \sup(A)\sup(B)$$

This is actually the same as proving that

for every pair of nonempty upper bounded sets of real numbers $U$ and $V$ we have $$\sup(U+V)=\sup U+\sup V$$

which is an easier exercise.

For an upper bounded (nonempty) set of positive numbers $X$, consider $X'=\{\log x:x\in X\}$ (natural logarithm). Then, due to the fact that logarithm and exponential are continuous, increasing and inverse of one another, $$ \sup X=\exp(\sup X') $$

Since $(AB)'=A'+B'$, we also have \begin{align} \sup(AB)&=\exp(\sup((AB)'))\\[4px] &=\exp(\sup(A'+B'))\\[4px] &=\exp(\sup A'+\sup B')\\[4px] &=\exp(\sup A')\exp(\sup B')\\[4px] &=(\sup A)(\sup B) \end{align}