What does exactly $ ∀ε>0$ means?

Question

I was proving a problem from textbook which states :
Let $a < b$ be real numbers and consider the set $T= ℚ ∩ [a,b]$. Then $supT=b$.
My attempt: Since $a < b$, so by density of $ ℚ $ there exists atleast one rational number between $a$ and $b.$ Thus $T ≠ ∅$. Moreover $b$ is an upper bound for $T$, therefore by axiom of completeness supT exists. Suppose $ε > 0$ be arbitrary. Then $b − ε < b.$ By density of rational numbers there exists $q ∈ℚ $ s.t $b − ε < q < b$.
Here I have to show that $q \in ℚ $ and i got stuck. Later I found a solution (online) :

I saw my solution is exactly similar. But after looking the next step I still didn't understand
How $r\in ℚ$ and $b- ε < r < b$ implies $r \in T$?
So my online friend told me "this is because of arbitrary small $ ε $. Moreover in real analysis we suppose $ ∀ ε > 0$ as arbitrary small $ ε > 0$. Even in your proof, we have $ 0 < ε < b -a $ ." (Notice this is exact his wording I am using all conclusion which I understood).

My questions:

1. what does $ ∀ ε > 0$ shows?
   Some people say this is arbitrary small number, that is $0 < ε <$ (some positive number, whatever your needed).
   But if so, then why we write " $ ∀ ε > 0$? Because this shows actually $ ∀x ∈ ℝ (x > 0)$, any positive real number not necessarily small.

2. How does "$ \cdots$ implies $r ∈ T$ " in above solution (pic)?

I have spend almost whole day to understand the concept but I didn't. I have found many posts on MSE about epsilon but none of them was answering my question.
Thank you.

Answer 1

In real analysis, many of the definitions are set around “for any epsilon greater than zero”. This ensures that our definition will hold for any positive number (often this is a notion of distance), this makes our definition more general than saying for any $1>\epsilon>0$, for example but most of the time we only need worry about small epsilon, so you could interpret this as epsilon being arbitrarily small. For the purposes of your proof you are using $b-a > \epsilon > 0$, since what you are actually showing is every interval is non-empty, so $b-\epsilon$ is not an upper bound for any such epsilon.

With respect to your proof, you seem to be making good progress, just remember the definition of supremum is least upper bound, perhaps an easier to understand proof strategy for this question would be to show $b$ is an upper bound, then get a contradiction by assuming some arbitrary number smaller than $b$ is an upper bound, since this is what is at the heart of your proof.