Understandig proof: Distance from Subset to Supremum 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.
 A: 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 $d(\sup S,S)=0$.
A: 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
$$d(\sup S, S)= 0.$$
