Following up with a previous question on $\sup(A)+\sup(B) = \sup(A + B)$

The question link is here:

Prove that $Sup(A + B) = Sup(A) + Sup(B)$

Can someone look at the answer given and explain why epsilon is introduced and how that whole second part works?

There is no need for $\epsilon$s.

The key idea is that if $x \le u$ for all $x \in X$ then $\sup X \le u$.

Note that if $a \in A, b \in B$ then $a+b \le \sup A + \sup B$ and so $\sup (A+B) \le \sup A + \sup B$.

Similarly, $a+b \le \sup (A+B) $, so $\sup A + b \le \sup (A+B)$ for all $b$ and hence $\sup A + \sup B \le \sup (A+B)$.

Because the definition of $\sup A$ ,for each $\epsilon > 0,\sup A < a+\epsilon.$

Remember $A$ is a subset of $\mathbb R$.