Supremum of a scalar multiple

Case $c>0$, in which $S$ is nonempty and bounded above: the least-upper-bound property supplies $u:=\sup S$, an upper bound of $S$ that is $\le$ every upper bound of $S$.

Case $c<0$, in which $S$ is nonempty and bounded below: $S$ has a greatest lower bound, and we set $\mathrm{\ell }:=\inf S$, a lower bound of $S$ with ${\mathrm{\ell }}^{\mathrm{\prime }}\le \mathrm{\ell }$ for every lower bound ${\mathrm{\ell }}^{\mathrm{\prime }}$ of $S$.

In the case $c>0$, every $s\in S$ satisfies $s\le u$, hence $sc\le uc$, that is $cs\le cu$; since the elements of $cS$ are exactly these $cs$ and $S\mathrm{\ne }\mathrm{\varnothing }$, the set $cS$ is nonempty and $cu$ is an upper bound of it.

In the case $c<0$, every $s\in S$ satisfies $\mathrm{\ell }\le s$, and multiplying by the negative $c$ reverses this to $sc\le \mathrm{\ell }c$, that is $cs\le c\mathrm{\ell }$; so $cS$ is nonempty and $c\mathrm{\ell }$ is an upper bound of it.

In the case $c>0$, let $\epsilon >0$ and put $\delta :=\epsilon c^{-1}$, so that $\delta c=\epsilon$; from $0\cdot c=0<\epsilon =\delta c$ and $c>0$ the equivalence form of [L2] gives $\delta >0$, so the epsilon characterisation applied to $S$ and $u$ yields $s\in S$ with $u-\delta <s$, and multiplying that inequality by $c>0$ gives $cu-\epsilon =(u-\delta )c<sc=cs$, an element of $cS$.

In the case $c<0$, let $\epsilon >0$ and put $\delta :=-\epsilon c^{-1}$, so that $\delta c=-\epsilon <0=0\cdot c$, which for the negative multiplier $c$ gives $\delta >0$ by [L2]; then $\mathrm{\ell }<\mathrm{\ell }+\delta$, and $\mathrm{\ell }+\delta$ cannot be a lower bound of $S$, since greatestness of $\mathrm{\ell }$ would force $\mathrm{\ell }+\delta \le \mathrm{\ell }$ and hence $\delta \le 0$; so some $s\in S$ fails $\mathrm{\ell }+\delta \le s$, which by trichotomy means $s<\mathrm{\ell }+\delta$, and multiplying by $c<0$ reverses it to $c\mathrm{\ell }-\epsilon =(\mathrm{\ell }+\delta )c<sc=cs$, an element of $cS$.

In the case $c>0$, the set $cS$ is nonempty and bounded above by $cu$, and for every $\epsilon >0$ some element of $cS$ exceeds $cu-\epsilon$, so $\sup (cS)$ exists and the epsilon characterisation identifies it: $\sup (cS)=cu=c\sup S$, which is claim 1.

In the case $c<0$, the set $cS$ is nonempty and bounded above by $c\mathrm{\ell }$, and for every $\epsilon >0$ some element of $cS$ exceeds $c\mathrm{\ell }-\epsilon$, so $\sup (cS)$ exists and equals $c\mathrm{\ell }=c\inf S$, which is claim 2.

A nonzero $c$ satisfies exactly one of $c>0$ and $c<0$, so the two cases are mutually exclusive and together exhaust the hypothesis $c\mathrm{\ne }0$, and each has been settled; both claims therefore hold.