In this proof:
https://proofwiki.org/wiki/Distance_from_Subset_to_Supremum
I don't understand why the two observations imply result.
Intuitively I get it, the distance is positive or zero and I can make it as near to zero as desired. But how to proper justify this?
Thanks.
On
If you want to justify it properly, you can do it like this. First, show that for any $a\in\mathbb{R}$, $$ a<\epsilon\text{ for all }\epsilon>0\implies a\leq0\tag{1} $$ You could equivalently show the contrapositive $$ a>0\implies\text{ there is an }\epsilon>0\text{ such that } a\geq\epsilon\tag{2} $$ Indeed, if $a>0$, then $a\geq a/2>0$ hence the choice $\epsilon:=a/2$ works. This shows $(2)$ and, equivalently, $(1)$.
Hence, in your case, $d(\sup S,S)\leq0$. Since you also have, as noted in the proof, $d(\sup S,S)\geq0$, you can conclude (by trichotomy) that $d(\sup S,S)=0$.
Maybe this can help you.
Prove: If $a<b+\epsilon$ for all $\epsilon>0$ then $a\leq b$.
Proof: Assume that $a<b+\epsilon$ for all $\epsilon>0$ and suppose $a>b$. We know that $a-b>0$, so that using our assumption, we get $a<b+(a-b)$. This means that $a<a$, which is not possible (because of the Trichotomy Law). So, we are forced to conclude that $a\leq b$ otherwise we will get a contradiction. QED
Applying this to the question, it has been shown that $$d(\sup S, S)<0+\epsilon$$ for all $\epsilon>0$. So, according to preceding result, we conclude that $$d(\sup S, S)\leq 0.\qquad (1)$$ But we know that by definition of the distance, $$d(\sup S, S)\geq 0.\qquad (2)$$ Combining $(1)$ and $(2)$, we get $$d(\sup S, S)= 0.$$