[proofplan]
Let $m = \inf A$ and let $B = \{\lambda a : a \in A\}$. Because multiplication by the negative number $\lambda$ reverses inequalities, the lower-bound property $m \le a$ for $A$ becomes the upper-bound property $\lambda a \le \lambda m$ for $B$. To prove leastness, take an arbitrary upper bound $u$ of $B$, divide by $\lambda < 0$ to obtain a lower bound $u/\lambda$ of $A$, and then use the greatest-lower-bound property of $m$.
[/proofplan]
[step:Define the scaled set and prove it is nonempty and bounded above]
Let $m = \inf A$, and define the scaled set
\begin{align*}
B = \{\lambda a : a \in A\} \subset \mathbb{R}.
\end{align*}
Since $A$ is nonempty, choose $a_0 \in A$. Then $\lambda a_0 \in B$, so $B$ is nonempty.
Since $m = \inf A$, the number $m$ is a lower bound for $A$. Thus, for every $a \in A$,
\begin{align*}
m \le a.
\end{align*}
Multiplying this inequality by $\lambda < 0$ reverses the inequality, giving
\begin{align*}
\lambda a \le \lambda m.
\end{align*}
Therefore every element of $B$ is at most $\lambda m$, so $\lambda m$ is an upper bound for $B$. Hence $B$ is nonempty and bounded above.
[guided]
We introduce two pieces of notation to separate the order argument from the set-builder notation. Let
\begin{align*}
m = \inf A,
\end{align*}
and define
\begin{align*}
B = \{\lambda a : a \in A\} \subset \mathbb{R}.
\end{align*}
First we verify that $B$ is a set for which a supremum can be discussed. Since $A$ is nonempty, there exists $a_0 \in A$. By the definition of $B$, the number $\lambda a_0$ belongs to $B$. Hence $B$ is nonempty.
Next we show that $B$ is bounded above. Because $m = \inf A$, the number $m$ is a lower bound for $A$. This means that for every $a \in A$,
\begin{align*}
m \le a.
\end{align*}
The scalar $\lambda$ is negative. Multiplication by a negative real number reverses the order, so the preceding inequality gives
\begin{align*}
\lambda a \le \lambda m.
\end{align*}
This holds for every $a \in A$. Equivalently, every element of the form $\lambda a$ with $a \in A$ is bounded above by $\lambda m$. Since those are exactly the elements of $B$, the number $\lambda m$ is an upper bound for $B$. Thus $B$ is nonempty and bounded above.
[/guided]
[/step]
[step:Show that every upper bound of the scaled set lies above $\lambda m$]
Let $u \in \mathbb{R}$ be an arbitrary upper bound for $B$. Then for every $a \in A$, the element $\lambda a$ belongs to $B$, so
\begin{align*}
\lambda a \le u.
\end{align*}
Dividing by $\lambda < 0$ reverses the inequality, and therefore
\begin{align*}
a \ge \frac{u}{\lambda}
\end{align*}
for every $a \in A$. Hence $u/\lambda$ is a lower bound for $A$.
Since $m = \inf A$ is the greatest lower bound of $A$, every lower bound of $A$ is at most $m$. Applying this to the lower bound $u/\lambda$, we obtain
\begin{align*}
\frac{u}{\lambda} \le m.
\end{align*}
Multiplying again by $\lambda < 0$ reverses the inequality, giving
\begin{align*}
u \ge \lambda m.
\end{align*}
Thus every upper bound $u$ of $B$ satisfies $\lambda m \le u$.
[guided]
To prove that $\lambda m$ is the least upper bound of $B$, it is not enough to know that $\lambda m$ is an upper bound. We must also prove that no smaller number can be an upper bound. So let $u \in \mathbb{R}$ be an arbitrary upper bound for $B$.
The statement that $u$ is an upper bound for $B$ means that every element of $B$ is at most $u$. For each $a \in A$, the number $\lambda a$ belongs to $B$, and therefore
\begin{align*}
\lambda a \le u.
\end{align*}
Now we want to translate this back into a statement about the original set $A$. Since $\lambda < 0$, dividing by $\lambda$ reverses the inequality. Hence, for every $a \in A$,
\begin{align*}
a \ge \frac{u}{\lambda}.
\end{align*}
This says exactly that $u/\lambda$ is a lower bound for $A$.
Because $m = \inf A$, the number $m$ is the greatest lower bound of $A$. Thus every lower bound of $A$ is less than or equal to $m$. Applying this to the lower bound $u/\lambda$, we get
\begin{align*}
\frac{u}{\lambda} \le m.
\end{align*}
Multiplying this inequality by the negative number $\lambda$ reverses the inequality once more, so
\begin{align*}
u \ge \lambda m.
\end{align*}
Therefore every upper bound $u$ of $B$ lies above $\lambda m$.
[/guided]
[/step]
[step:Conclude that $\lambda m$ is the supremum of the scaled set]
From the first step, $\lambda m$ is an upper bound for $B$. From the second step, every upper bound $u$ of $B$ satisfies $\lambda m \le u$. Therefore $\lambda m$ is the least upper bound of $B$. Hence
\begin{align*}
\sup B = \lambda m.
\end{align*}
Substituting $B = \{\lambda a : a \in A\}$ and $m = \inf A$ gives
\begin{align*}
\sup\{\lambda a : a \in A\} = \lambda \inf A.
\end{align*}
This proves the theorem.
[/step]
