[proofplan]
Let $s = \sup A$. We prove each identity directly from the least-upper-bound property: first show the proposed number is an upper bound of the transformed set, then show it is no larger than any other upper bound. Translation uses order compatibility with addition, while positive scaling uses order compatibility with multiplication by the positive number $\lambda$ and division by $\lambda$.
[/proofplan]
[step:Fix the supremum of the original set]
Since $A$ is nonempty and bounded above, the supremum $s := \sup A$ exists in $\mathbb{R}$. By definition of supremum, $s$ is an upper bound for $A$, so for every $a \in A$,
\begin{align*}
a \le s.
\end{align*}
Also, if $w \in \mathbb{R}$ is any upper bound for $A$, then
\begin{align*}
s \le w.
\end{align*}
[/step]
[step:Translate upper bounds by adding $c$]
Define the translated set $T \subset \mathbb{R}$ by
\begin{align*}
T := \{a + c : a \in A\}.
\end{align*}
Because $A$ is nonempty, choose $a_0 \in A$; then $a_0 + c \in T$, so $T$ is nonempty. For any $x \in T$, there exists $a \in A$ such that $x = a + c$. Since $a \le s$, adding $c$ to both sides gives
\begin{align*}
x = a + c \le s + c.
\end{align*}
Thus $s + c$ is an upper bound for $T$, and in particular $T$ is bounded above.
Now let $u \in \mathbb{R}$ be any upper bound for $T$. For every $a \in A$, the element $a + c$ lies in $T$, so
\begin{align*}
a + c \le u.
\end{align*}
Subtracting $c$ from both sides gives
\begin{align*}
a \le u - c.
\end{align*}
Hence $u - c$ is an upper bound for $A$. Since $s = \sup A$, the least-upper-bound property gives
\begin{align*}
s \le u - c.
\end{align*}
Adding $c$ to both sides gives
\begin{align*}
s + c \le u.
\end{align*}
Therefore $s + c$ is an upper bound for $T$ that is less than or equal to every upper bound of $T$. Hence
\begin{align*}
\sup T = s + c.
\end{align*}
Equivalently,
\begin{align*}
\sup\{a + c : a \in A\} = \sup A + c.
\end{align*}
[guided]
Define the translated set $T \subset \mathbb{R}$ by
\begin{align*}
T := \{a + c : a \in A\}.
\end{align*}
We must prove that $s + c$ is the least upper bound of $T$. The proof has two parts: it must be an upper bound, and it must be less than or equal to every other upper bound.
First, $T$ is nonempty because $A$ is nonempty. Choose $a_0 \in A$; then $a_0 + c \in T$. To see that $s + c$ is an upper bound, take an arbitrary $x \in T$. By the definition of $T$, there exists $a \in A$ such that $x = a + c$. Since $s = \sup A$, the number $s$ is an upper bound for $A$, so $a \le s$. Addition preserves order in $\mathbb{R}$, hence
\begin{align*}
x = a + c \le s + c.
\end{align*}
Because this holds for every $x \in T$, the number $s + c$ is an upper bound for $T$.
Now we prove minimality. Let $u \in \mathbb{R}$ be an arbitrary upper bound for $T$. Since $a + c \in T$ for every $a \in A$, the upper-bound property of $u$ gives
\begin{align*}
a + c \le u
\end{align*}
for every $a \in A$. Subtracting $c$ from both sides gives
\begin{align*}
a \le u - c
\end{align*}
for every $a \in A$. Therefore $u - c$ is an upper bound for $A$. Since $s$ is the least upper bound of $A$, we have
\begin{align*}
s \le u - c.
\end{align*}
Adding $c$ to both sides yields
\begin{align*}
s + c \le u.
\end{align*}
Since $u$ was an arbitrary upper bound of $T$, every upper bound of $T$ is at least $s + c$. Together with the fact that $s + c$ is itself an upper bound, this proves
\begin{align*}
\sup T = s + c.
\end{align*}
That is,
\begin{align*}
\sup\{a + c : a \in A\} = \sup A + c.
\end{align*}
[/guided]
[/step]
[step:Scale upper bounds by the positive number $\lambda$]
Define the positively scaled set $S \subset \mathbb{R}$ by
\begin{align*}
S := \{\lambda a : a \in A\}.
\end{align*}
Because $A$ is nonempty, choose $a_1 \in A$; then $\lambda a_1 \in S$, so $S$ is nonempty. For any $y \in S$, there exists $a \in A$ such that $y = \lambda a$. Since $a \le s$ and $\lambda > 0$, multiplication by $\lambda$ preserves order, so
\begin{align*}
y = \lambda a \le \lambda s.
\end{align*}
Thus $\lambda s$ is an upper bound for $S$, and in particular $S$ is bounded above.
Now let $v \in \mathbb{R}$ be any upper bound for $S$. For every $a \in A$, the element $\lambda a$ lies in $S$, so
\begin{align*}
\lambda a \le v.
\end{align*}
Since $\lambda > 0$, division by $\lambda$ preserves order, and therefore
\begin{align*}
a \le \frac{v}{\lambda}.
\end{align*}
Hence $v/\lambda$ is an upper bound for $A$. Since $s = \sup A$, the least-upper-bound property gives
\begin{align*}
s \le \frac{v}{\lambda}.
\end{align*}
Multiplying both sides by $\lambda > 0$ gives
\begin{align*}
\lambda s \le v.
\end{align*}
Therefore $\lambda s$ is an upper bound for $S$ that is less than or equal to every upper bound of $S$. Hence
\begin{align*}
\sup S = \lambda s.
\end{align*}
Equivalently,
\begin{align*}
\sup\{\lambda a : a \in A\} = \lambda \sup A.
\end{align*}
[/step]
