This is just alternative proof that uses lemma 1.3.8 to the proof in here. The first part of the proof is exactly the same but proving the second part is different.

Given sets A and B, define the set $A + B = \{a + b : a \in A \text{ and } b \in B\}$. Prove that $\sup A + B = \sup A + \sup B $

For the definitions of an upper bound and the least upper bound of a set. See This.

Proof:

Let $A$ and $B$ be sets and let $A + B$ be the set defined above. We want to prove that $\sup A+B = \sup A + \sup B$. To do this, we will prove:

$\sup (A+B) \leq \sup A + \sup B$.
$\sup (A+B) \geq \sup A + \sup B$.

From these two inequalities we will then conclude that $\sup A+B = \sup A + \sup B$. To prove that $\sup A+B \leq \sup A + \sup B$, let $a$ be an arbitrary element in $A$. Then we must have,

$$ \begin{align*} a &\leq \sup A. \end{align*} $$

Similarly, Let $b$ be an arbitrary element in $B$. Then we must have,

$$ \begin{align*} b &\leq \sup B. \end{align*} $$

Adding both inequalities will result in

$$ \begin{align*} a + b &\leq \sup A + \sup B. \end{align*} $$

Since $a$ and $b$ were arbitrary, this means that $\sup A + \sup B$ is an upper bound for $A + B$. But we know the least upper bound for $A+B$ is $\sup A + B$ and by the definition of the least upper bound $\sup A + B$ must be less than or equal to any other bound on $A + B$. Therefore we can write,

$$ \begin{align*} \sup (A+B) &\leq \sup A + \sup B. \end{align*} $$

And this concludes the first part of the proof. To prove condition two, define an arbitrary $\epsilon > 0$. By lemma 1.3.8 we know that there exists some element $x \in A$ such that,

$$ \begin{align*} x > \sup A - \epsilon/2. \end{align*} $$

(Side note: the lemma states that $\sup A$ is a least upper bound iff for any $\epsilon > 0$, there exists some element in $A$ such that $a > \sup - \epsilon$ so really any tiny epsilon taken away from $\sup A$ will render it not a least upper bound anymore)
Similarly, there exists an element $y \in B$ such that,

$$ \begin{align*} y > \sup B - \epsilon/2. \end{align*} $$

Adding both inequalities we see,

$$ \begin{align*} x + y > \sup A + \sup B - \epsilon. \end{align*} $$

But from part 1 of the proof we know that $\sup A + \sup B$ is an upper bound for $A+B$ where we saw that if for any arbitrary elements $a \in A$ and $b \in B$, we have

$$ \begin{align*} a + b \leq \sup A + \sup B. \end{align*} $$

So we have two things. First $\sup A + \sup B$ is an upper bound on $A+B$. Second, we proved above that for any $\epsilon > 0$, there exists some elements $x \in A$ and $y \in B$ such that $x + y > \sup A + \sup B - \epsilon$. So by lemma 1.3.8, $\sup A + \sup B$ is a least upper bound for $A + B$ and so we must have,

$$ \begin{align*} \sup A + \sup B \leq \sup (A + B). \end{align*} $$

as required. $\blacksquare$

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