You should understand that many of the steps you've described as "fey and prescient" were probably one of several attempts made while trying to sketch out a proof. The presentation of mathematics will almost always strip away failed attempts, or re-organize what was at first simply scratch work. Admittedly, those comfortable with material will often be able to see the best path in a way that is a bit eerie, but let's look at this problem, and how one might get to this proof in a way that seems a bit less prescient.
A very useful technique when sketching a proof is to try and work backwards, or to reduce your problem to an equivalent one that you can grasp. Let's start with the second one, which is "easier", the following inequality:
$$\sup(A + B) \leq \sup(A) + \sup(B)$$
We ask ourselves what this really means. What do we have to do to prove it? The left-hand side is the supremum of $A+B$. To prove a number is greater than or equal to a supremum, we just need to show it is an upper bound of the same set. This comes directly from the definition of a supremum. So we just want to show that for any $a+b \in A+B$, we have $$a+b \leq \sup(A) + \sup(B)$$
This is clearly true, as $\sup(A) \geq a$, and $\sup(B) \geq b$. So we are done, $\sup(A) + \sup(B)$ is an upper bound on $A+B$, making it larger than the supremum thereof. Clean this up, remove the exposition, and you suddenly have the seemingly prescient short proof.
What does this tell us about the other direction? Namely the inequality
$$\sup(A + B) \geq \sup(A) + \sup(B)$$
Well, we've seen from our previous proof that it's nice to work with things that are bigger than the supremum of a single set. So we try and isolate $\sup(A)$ on the right. We get the equivalent inequality:
$$\sup(A + B)-\sup(B) \geq \sup(A)$$
Note: we haven't proven this, but if we can prove it, we'll be done. Now again, something being bigger than a supremum just means they are an upper bound of the set: so we just need to prove that for any $a\in A$
$$\sup(A + B)-\sup(B) \geq a$$
Again though, we like working with things bigger than a supremum, so we can again rearrange to get the equivalent condition:For any $a \in A$
$$\sup(A + B)- a \geq \sup(B)$$
And now we use our favorite fact, that this is true if and only if for all $a \in A$ we have that $\sup(A+B) - a$ is an upper bound on $B$. Written better, we've shown our original inequality equivalent to: For all $a \in A$, for all $b \in B$:
$$\sup(A + B)- a \geq b$$
or equivalently:For all $a \in A$, for all $b \in B$:
$$\sup(A + B)\geq a + b$$
Which is trivially true. So we reverse the implications, write it up in order, remove our scratchwork, and tada: "fey and prescient" proof!
Note 1: When choosing how to rearrange our inequality, we always tried to get a positive supremum term alone on the right. $A$ and $B$ are indistinguishable in their roles in the original claim, so it didn't matter which we isolated first. Note also this goal barred us from ever moving around $\sup(A+B)$ as it would have appeared as a negative, unhelpful to us.
Note 2: Though it worked well enough here, beware of relying to heavily on backwards work. You have to make sure your implications are going the right way. Here we used only equivalencies.
Hope this helps. I'd advise you try this on a delta-epsilon type problem as well, such as proving the quotient law for limits, or the continuity of a polynomial from first principles.
