To prove that $\sup A + \sup B \leq \sup(A+B)$ it suffices to prove that $\forall \epsilon>0, \sup A + \sup B \leq \sup(A+B) + \epsilon$? 
To prove that $$\sup A + \sup B \leq \sup(A+B)$$ it suffices to prove
that $$\forall \epsilon>0, \sup A + \sup B \leq \sup(A+B) + \epsilon$$

How do we know (prove) that this is true?
Note: $\sup A$ is the supremum (least upper bound) of a set $A$.
 A: Suppose that the former statement is false: then $\sup (A+B)<\sup A + \sup B$, so $\exists \epsilon > 0: \sup(A+B)+\epsilon<\sup A + \sup B$ (for instance, $\epsilon := \frac{\sup A + \sup B-\sup (A+B)}{2}$ will work).
But then $\forall \epsilon > 0: \sup (A+B)+\epsilon \ge \sup A + \sup B$ is false.
(Because the claim is that the latter statement "suffices", we only need to show that it implies the former statement, which we have done by contrapositive, and not the reverse implication too.)
Notice that this does not actually use any specific property of $\sup$, other than that $\sup(S)$ is a real number.
A: In fact $$\sup A+\sup B=\sup\{A+B\}$$
Denote $a=\sup A$, $b=\sup B$. Then $\forall x\in A, y\in B$, we have $x+y\le a+b$, which implies $$\sup\{A+B\}\le a+b$$
On the other hand, $\forall\varepsilon>0$, $\exists x'\in A, y'\in B$, such that $x'>a-\varepsilon, y'>b-\varepsilon$, then $x'+y'>a+b-2\varepsilon$, which implies $$\sup\{A+B\}>a+b-2\varepsilon$$
Thus $$\sup\{A+B\}=a+b=\sup A+\sup B$$
A: In fact, if you look in depth at your question, interestingly it has nothing to do with $\sup A, \sup B, \sup(A+B)$. What you're ultimately asking is

To prove that a number $a \in \mathbb R$ is non-positive ($a \le 0$), it is sufficient to prove that for any $\epsilon \gt 0$, we have $a \le \epsilon$.

This is indeed true by contradiction as if for some $\epsilon_0 \gt 0$, you would have $a \gt \epsilon_0$, then $a \gt \epsilon_0 \gt 0$ proving that $a$ is strictly positive.
That being said, just apply this result in your particular case to
$$a = \sup A + \sup B - \sup(A+B).$$
You'll get $a \le 0$ as desired.
