I know there are other threads in this site that gives alternate proofs to this. However, I was hoping to see if you thought my approach (which I don't see on this site yet) is valid. The proof follows:
$\exists \ \bar{a} \in A \ \text{ s.t. } \ \forall a \in A, \ \bar{a} \geq a$ and $\ \bar{a} \leq \sup (A)$. Similarly, $\exists \ \bar{b} \in B \ \text{ s.t. } \ \forall b \in B, \bar{b} \geq b$ and $\bar{b} \leq \sup(B)$. It follows that $\exists \ \bar{c}=\bar{a}+\bar{b} \in A+B \ \text{ s.t. } \ \forall c \in A+B, \ \bar{c} \geq c $ and $\bar{c} \leq \sup(A) + \sup (B)$. This establishes that $\sup (A) + \sup (B)$ is an upper bound for $A+B$. We also know that if $a' < \sup (A), \exists \ a \in A \text{ s.t. } a>a'$. Similarly, if $ b' < \sup(B), \ \exists \ b \in B \text{ s.t. } b>b'$ $\therefore \exists \ c \in A+B \text{ s.t. } c > \sup(A)+b', \ c>a'+\sup(B), \ \text{ and } c>a'+b' \implies \sup (A+B) = \sup(A)+\sup(B)$.
The first line itself is incorrect in two places.
First of all, you are saying that there is an $\bar a \in A$ such that for all $a \in A$, $\bar a \geq a$. This is incorrect, there are umpteen counterexamples, some given above. Yes, the $\bar a$ may be there, but there's no guarantee it will be in $A$ itself.
Furthermore, by what you have given, $\bar a$ is an upper bound of the set $A$, and the supremum of $A$ is the least upper bound, so $\sup A \leq \bar a$ should follow, and not the other way. The same two wrongs have been made for $b$, and they certainly don't make a right.
What you are getting correct is the idea : you first want to prove that $\sup A + \sup B$ is an upper bound , then that it is the least upper bound.
To show that it is an upper bound, it's enough to show for all $a \in A$, $b \in B$, that $a + b \leq \sup A + \sup B$. But then, $\sup A \geq a$ and $\sup B \geq b$, so this follows straight away.
From here, you need to show that it is indeed the least upper bound. This procedure will tell you how to do it.You can keep it in mind if you encounter a similar question.
Let $\epsilon > 0$. As $\sup A - \frac\epsilon 2$ is not an upper bound of $A$, there is an element $a_0 \in A > \sup A - \frac\epsilon 2$.
Similarly, there is an element $b_0 > \sup B - \frac\epsilon 2$.
Adding these up, $a_0 + b_0 > \sup A + \sup B - \epsilon$.
So, for all $\epsilon > 0$, $\sup A + \sup B - \epsilon$ is not an upper bound of $A+B$. So the least upper bound is at least $\sup A + \sup B$, but we've already seen this is an upper bound, so it follows it's the least upper bound.
Hence, the lemma is proven.
Now, you can use this procedure to try and prove that the supremum of the set $\{a \times b : a \in A, b \in B\}$ is $\sup A \sup B$, for $A, B$ subsets of positive real numbers.