In terms of the logic of the proof, what is the difference between $a+b$ and $A+B$ in the proof that $\sup(A+B) = \sup A + \sup B$? Of course, $a + b$ is an element of $A+B$, but what I'm unable to wrap my head around, is when does $a+b$ behave like if it was the entirety of $A+B$, and when does it behave as a single element of $A+B$?
This is in the context of trying to prove $\sup A + \sup B \leq \sup(A+B)$. For instance, I'm confused about the difference in these statements:

*

*Let $a,b$ be elements of the subintervals $A,B$ of the reals, and let $x = a + b$. Then $x \leq \sup A + \sup B$.

*Let $a,b$ be elements of the subintervals $A,B$ of the reals, and let $x \in A+B$. Then $x \leq \sup(A + B)$.

And also the difference between these: (I'm also not sure, can we actually instead infer that $\sup A + \sup B$ is a least upper bound from $a \leq \sup A$ and $b \leq \sup B$)?


*Since $a \leq \sup A$ and $b \leq \sup B$, we have that $\sup A + \sup B$ is an upper bound of $a + b$.

*Since $a \leq \sup A$ and $b \leq \sup B$, we have that $\sup A + \sup B$ is an upper bound of $A + B$.

Any help or reading suggestions is very welcome!

You asked "When does $a+b$ behave like if it was the entirety of $A+B$."  I would say it never does; it is always a single element of $A+B$.  However, if $a$ and $b$ are introduced into the proof as arbitrary elements of $A$ and $B$, and you then prove some statement about them (treating them as unspecified single elements of $A$ and $B$), then you can say that since they were arbitrary, the statement is true for all $a \in A$ and $b \in B$.
Your statement 3 is written in a confusing way.  It would be clearer to write: Since $a \le \sup A$ and $b \le \sup B$, we have that $a+b \le \sup A + \sup B$.
Your statement 4 is correct, although I think it would be clearer to write it like this:  Let $x$ be an arbitrary element of $A+B$.  Then there are $a \in A$ and $b \in B$ such that $x = a+b$.  Since $a \le \sup A$ and $b \le \sup B$, $x = a+b \le \sup A + \sup B$.  Since $x$ was arbitrary, we can conclude that for all $x \in A+B$, $x \le \sup A + \sup B$, so $\sup A + \sup B$ is an upper bound of $A+B$.  This proves that $\sup(A+B) \le \sup A + \sup B$.
