(1) Yes, the intended order is the usual one on $\mathbb{R}$.
(2) As JavaMan noted, $\max A$ need not exist: what if $A = [0,1)$, for instance? Let’s look at a concrete example in which your argument fails completely. Take $A = [0,1)$ and $B=[2,3)$; what numbers are in $C$? You should be able to convince yourself fairly readily that $C=[2,4)$. In this example none of the three sets has a maximum element, though all are bounded above.
To show that $\sup C=\sup A+\sup B$, you need to begin by showing that $\sup C$ exists, i.e., that $C$ is bounded above. Since you want $\sup A+\sup B$ to be the supremum of $C$, you might try to show first that $\sup A+\sup B$ is an upper bound for $C$, i.e., that $a+b\le \sup A + \sup B$ whenever $a\in A$ and $b\in B$. Having done this, you’d know both that $\sup C$ exists and that $\sup C \le \sup A + \sup B$.
The natural next step would be to try to show that $\sup A + \sup B \le \sup C$, so as to be able to conclude that $\sup C = \sup A + \sup B$. It’s a little hard to see right away just what it means for $\sup A + \sup B$ to be less than or equal to $\sup C$, so try assuming the opposite and getting a contradiction: what happens if $\sup C < \sup A + \sup B$? You’d get a contradiction if you could show that whenever $u<\sup A + \sup B$, there are $a\in A$ and $b\in B$ such that $a+b>u$. This is where the real work in this argument takes place.
HINT: Let $d=(\sup A + \sup B)-u$, and look at the intervals $\left(\sup A-\frac{d}2,\sup A\right)$ and $\left(\sup B-\frac{d}2,\sup B\right)$.
