If sup A < sup B, there exists an element b ∈ B that's an upper bound for A. (S.A. pp 18 q1.3.8) 
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?

 A: 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!
A: Your figure gives the idea behind the proof. Clearly we must have elements in $B$ which are as near to $B$ as we please. The second arc in the figure just near $\sup B$ represents one such element on the number line. In order that this element (name it $b$) also be an upper bound for $A$, it is sufficient to make it greater than or equal to $\sup A$. Thus we need an element $b$ near $\sup B$ and greater than or equal to $\sup A$. Since $\sup A < \sup B$ it is possible if we can choose $b$ nearer to $\sup B$ than $\sup A$ is near to $\sup B$. Thus the distance between $\sup B$ and $b$ (i.e. $\sup B - b$) should be less than the distance between $\sup A$ and $\sup B$ (i.e. $\sup B - \sup A$). Thus we take $\epsilon = \sup B - \sup A$ and try to find $b$ such that $\sup B - b < \epsilon$.
