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 $\ell := \inf S$, a lower bound of $S$ with $\ell' \le \ell$ for every lower bound $\ell'$ 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 $S \ne \emptyset$, the set $cS$ is nonempty and $cu$ is an upper bound of it.

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

In the case $c > 0$, let $\varepsilon > 0$ and put $\delta := \varepsilon c^{-1}$, so that $\delta c = \varepsilon$; from $0 \cdot c = 0 < \varepsilon = \delta c$ 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 $cu - \varepsilon = (u - \delta)c < sc = cs$, an element of $cS$.

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

In the case $c > 0$, the set $cS$ is nonempty and bounded above by $cu$, and for every $\varepsilon > 0$ some element of $cS$ exceeds $cu - \varepsilon$, 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 $c\ell$, and for every $\varepsilon > 0$ some element of $cS$ exceeds $c\ell - \varepsilon$, so $\sup(cS)$ exists and equals $c\ell = c \inf S$, 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 $c \ne 0$, and each has been settled; both claims therefore hold.