$a\in A$, $b\in B$, $\alpha=\sup A$, $\beta=\sup B$, $C=\{a+b:a\in A, b\in B\}$, $\gamma$ is an upperbound for C.
I am trying to show that $\alpha+\beta\le\gamma$ i.e. $\sup C = \sup A + \sup B$.
I have determined that $b\le \beta \le \gamma - a$ and that $a\le \alpha \le \gamma - b$
I combined these inequalities to $a+b \le \alpha + \beta \le 2\gamma - (a+ b)$.
So $2(a+b) \le \alpha + \beta + (a+b) \le 2\gamma$
Clearly $\alpha + \beta + (a+b) \le 2(\alpha + \beta)$, but how do I know that $ 2(\alpha + \beta) \le 2\gamma$? Once I show this I divide by 2 and my proof is done, but I am stuck at this point.
Let us fix an arbitrary $b \in B$. Then for every $a \in A$ we have \begin{equation} a + b \leq \gamma. \end{equation} This is true for every $a \in A$. Thus it is also true for $\alpha$ (to see this, choose a sequence of elements in $A$ which converge to $\alpha$.) That is, \begin{equation} \alpha + b \leq \gamma. \end{equation} Now, this inequality holds for every $b \in B$ and thus it also holds for $\beta$.