My Figure:
By definition of $\sup B$, $\sup B$ is an upper bound for $B$.
Set $e = \sup B − \sup A > 0$.
By Lemma 1.3.7, there exists an element $b ∈ B$ satisfying $\begin{align} & \sup B − e < b \\ & = \sup A \end{align}$,
Because $\sup A$ is an upper bound for $A$, then $b$ is an upper bound for $A$ as well.
1. What's the modus operandi of the proof please? I'm not questioning for proofs.
I know each sentence is warranted. I don't grasp the nub of the proof?2. How can you presage to set $e = \sup B − \sup A$?
Then I see everything after behaves, but this feels eerie and fey.3. Apart from my figure, any other intuition please?
The intuition behind the proof is this. The strict inequality first of all is very important. The result need not be true if the condition is $\sup A \le \sup B$. If $\sup A \lt \sup B$ what you need to see is there are real numbers (uncountably many in fact) between $\sup A$ and $\sup B$. Why because $r = (\sup B - \sup A ) \gt 0$ and forms an interval on the line. Think what happens if none of these real numbers are in $B$. Then none of the elements that are greater than $\sup A$ are in $B$. That is there are no elements in $B$ which are greater than $\sup A$. Then $\sup A$ is an upper bound for $B$. Can this be possible. NO!! Why? Since $\sup B$ is the least upper bound of $B$ and $\sup A$ is a value less than it. This is the story in layman's terms.
I should add what the above contradiction proves. It says there are numbers in $B$ which are greater than $\sup A$ and hence are an upper bound for $A$.
Here is a Lemma that I have always found useful in understanding the intuition behind suprema. $u = \sup A \iff x \in A \implies x \le u $ and $b \lt u \implies \exists \ x' \in A$ such that $b \lt x'$. This particular Lemma makes a lot of proofs simple.
A direct application for your question states that $\exists b \in B$ such that $ b \gt \sup A$. End of Story!