let $\displaystyle S$ be nonempty and bounded above, then $\displaystyle T$ is also nonempty and bounded above, so both suprema exist. Let's denote $\displaystyle q = \mbox{sup }S$.
Is $\displaystyle aq$ an upper bound of $\displaystyle T$? Yes, because any $\displaystyle t \in T$ can be written as $\displaystyle t = as$ for some $\displaystyle s \in S$ and we have $\displaystyle s\leq q$ and $\displaystyle a$ is positive, so $\displaystyle t=as \leq aq$.
Is $\displaystyle aq$ the least upper bound? If it were not, there would be some $\displaystyle p \in \mathbb{R}$ such that $\displaystyle p<aq$ and $\displaystyle p$ is an upper bound of $\displaystyle T$. Then $\displaystyle p/a$ is an upper bound of $\displaystyle S$, because for any $\displaystyle s\in S$ we have $\displaystyle as \in T$ so $\displaystyle as \leq p$ and $\displaystyle s\leq p/a$. Furthermore, $\displaystyle p/a < q$. But this would mean that $\displaystyle q$ is not the least upper bound of $\displaystyle S$, which is contradiction.
We conclude that $\displaystyle aq = \mbox{sup }T$.