Proof verification of $a\sup S=\sup(aS)$ I'm trying to show that $a\sup S=\sup aS$ where $a>0$ and $aS:=\{as:s\in S\}$. 
My attempt:
Let $u$ be an arbitary upper bound for the set $S$ then by definition we have that $\forall s\in S\Rightarrow s\leq u$ so $as\leq au$ so $au$ is clearly an upper bound for the set $aS$. To show that this is the supremum of $aS$ by the archimedean property we have that $\forall s\in S$ and $\varepsilon_0>0\Rightarrow |s-\sup S|<\varepsilon_0$, multiplying both sides by $a>0$ and taking $\varepsilon_0=\frac{\varepsilon}{a}>0$ we obtain $$a|s-\sup S|=|as-a\sup S|<a\cdot\varepsilon_0=a\cdot\frac{\varepsilon}{a}=\varepsilon$$
This shows that $a\sup S=\sup(aS)$

Your proof is essentially correct, but I would write it a bit differently.
Let $u=\sup S$. Then, for all $s\in S$, $s\le u$, which implies $as\le au$. Therefore $au\ge\sup aS$, because it is an upper bound for $aS$.
Let now $\varepsilon>0$. We want to find $t\in aS$ such that $t>au-\varepsilon$. Since $u=\sup S$, there exists $s\in S$ with $s>u-\frac{\varepsilon}{a}$; then
$$
as>au-\varepsilon
$$
so $t=as\in aS$ is an element with the required property.
