Solutions

We have that $a\le\sup A$ for all $a\in A$ and $b\le\sup B$ for all $b\in B$. This means that

$$a+b\le\sup A+\sup B$$

for all $a\in A$ and $b\in B$. Hence, $\sup A+\sup B$ is an upper bound of $A+B$. Consequently, by definition,

$$\sup(A+B)\le\sup A+\sup B.$$

Now, by definition there exist $a\in A$ and $b\in B$ such that

$$\sup A-a<\frac{\varepsilon}{2}$$

and

$$\sup B-b<\frac{\varepsilon}{2}$$

for any $\varepsilon>0$. This implies that

$$\sup A+\sup B<a+b+\varepsilon\le\sup(A+B)+\varepsilon.$$