How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$
Set $x=\sup A$, $y=\sup B$. Given $a\in A, b\in B$ we have $a\le x, b\le y$ so $a+b\le x+y$ so it's an upper bound. Now take $\varepsilon > 0$ and find $a,b$ such that $a>x-\varepsilon/2, b>y-\varepsilon/2$ and you have $a+b > x+y - \varepsilon$. That suffices (since it means that every potential "smaller upper bound" $x+y-\varepsilon$ is not really an upper bound).
Show that $\sup(A+B)$ is less than or equal to $\sup(A)+\sup(B)$ by showing that the latter is an upper bound for $A+B$. Then show that $\sup(A)+\sup(B)$ is less than or equal to $\sup(A+B)$ by showing that $\sup(A+B)$ is an upper bound for $A+\sup(B)$ and that $\sup(A+\sup(B)) = \sup(A)+\sup(B)$.
The following proofs are based on Question 2 here. The nub is to prove two inequalities: $$\sup(A+B) \geq \sup A+ \sup B \tag{1}$$
$$\sup(A + B) \leq \sup A + \sup B \tag{2}$$
Proof of $(2)$:
By definition of $A + B$ and $\sup(A + B)$, for all $a \in A$ and $b \in B$,
$$\color{seagreen}{a + b} \color{seagreen}{\leq \sup (A + B)}.$$ Subtract $b$ from both sides:
$$a = \color{seagreen}{a + b} - b \color{seagreen}{\leq \sup (A + B)} - b.$$
Hence, if we fix $b \in B$, then $\color{seagreen}{\sup (A + B)} - b$ is an upper bound for $$\color{seagreen}{A + B} - B = A.$$
And so by definition of $\sup A$, for every $b \in B$, $$\sup A \leq \sup (A+ B) − b.$$ Rearrange the previous inequality: $\color{magenta}{b} \leq \sup(A +B) − \sup A$ for all $b \in B$.
Hence, $\sup(A +B) − \sup A$ is an upper bound for any $\color{magenta}{b}$.
By the definition of supremum, the previous inequality means: $$\color{magenta}{\sup B} \leq \sup(A + B) − \sup A \iff \sup A + \sup B \leq \sup(A + B).$$
Proof of $(1)$:
Since $\sup A$ is an upper bound for $A$, $a \leq \sup A$ for all $a \in A$. Then $b \leq \sup B$ for all $b \in B$. Hence $a + b \leq \sup A + \sup B$ for all $x \in A$ and $y \in B$. Hence, $\sup A + \sup B$ is an upper bound for $A + B$. Hence, by definition of supremum, $\sup A + \sup B \geq \sup(A + B)$.