This lecture note proved $(rs)^{t}=r^{t}s^{t}$ for a real number $t\gt 0$ by using the following statement: $$\text{sup}(AB)=\text{sup}A\,\text{sup}B.$$ I found this specific step problematic
For the property, if $r,s\gt 1$, then we can just use equation (1), noting that $\{(rs)^{a}:a\in\mathbb{Q}\,\text{and}\,a\le t\}=AB$, where $A=\{r^{a}:a\in\mathbb{Q}\,\text{and}\,a\le t\}$ and $B=\{s^{a}:a\in\mathbb{Q}\,\text{and}\,a\le t\}$.
because $AB$ is defined as $\{ab:a\in A,b\in B\}.$ In other words, there are some elements in $AB$ where their exponents do not match, thereby not satisfying the above beautiful representation $(rs)^{a}.$ In conclusion, my question is
Is my hypothesis right, which means the above proof does not hold?
You are absolutely right: in fact, $$ AB \ne \{(rs)^a \mid a \in \mathbb{Q}, a \le t \} $$ because of the problem you mention. However, we can salvage the proof. Let $C = \{(rs)^a \mid a \in \mathbb{Q}, a \le t \}$, and note that $$ C \subseteq AB $$ and moreover, for any element $r^a s^b \in AB$, $r^a s^b \le r^{\max(a,b)} s^{\max(a,b)} \in C$.
This implies that even though the two sets are not the same, they have the same supremum. So then the proof goes through because $$ (rs)^t = \sup C = \sup AB = \sup A \sup B = r^t s^t. $$