I might answer later but for now let me just say there is a minor typo. "... exists a ball $K$' that is concentric with $K$, has radius greater than that of $\color{red}{K}$, and is disjoint from $S$" is not quite right. What (I think) they mean is: ".... greater than that of $\color{blue}{B}$...". A ball centred at $K$'s centre with radius $\ge$ the radius of $K$ necessarily contains $x$ so is not disjoint from $S$

Thanks for replying, @FShrike. This would make a lot more sense, however, the author uses the version above for proving the proposition I first stated. I can add the proof of the proposition too, if that makes it more clear what is going on.

Maybe sketching how this exercise is used in the proof of the main proposition would be useful, perhaps the typo carried over

@FShrike I added the full proof. Not expecting you to read this, but it all kind of makes sense until the very end, when the author uses the above exercise to invoke a contradiction. Although not stated anywhere, I assumed in the proof that $X$ is $\mathbb R^N$.

Hmm, maybe it's the "concentric" claim that was a typo! They definitely do need a larger radius here. Having read through that "proof" though, I must remark - they nowhere seem to use the definition of $z$ as some point on a line $a\to b$ with $a,b\in S$. This is obviously going to be a crucial part of any correct argument... otherwise they'll just be proving that $S=\Bbb R^N$ (the only property of $z$ seemingly used so far is that $z\notin S$). So, I'm suspicious of this proof

That could indeed be another possibility. After all, $K_n$ will always be shifted from $K$ slightly, so I do not see how it could possibly be concentric.