how show that $b$ is the supremum of an interval $[a,b)$ I think it is obvious that the supremum of an interval of the form $[a,b)$ is $b$ but I am trying to show it myself.
My try:
$b$ is an upper bound, if it is a supremum, then $\exists x \in [a,b)$ s.t. $x+\epsilon>b\geq x$ for some $\epsilon>0$ (epsilon characterization of the supremum)
Assume it is not, then $\forall x \in [a,b)$, $ x+\epsilon\leq b <x$ but this means that $b$ isn't an upper bound, contradiction. Thus, $b$ must be the supremum of the interval. (also, one can notice that the inequality would imply that $\epsilon<0$ which is another contradiction)
Any help in verifying my error, correcting it, or suggesting better ways to prove this is appreciated.
 A: Your proof is confusingly written.
Problem 1:

if it is a supremum, then $\exists x \in [a,b)$ s.t. $x+\epsilon>b\geq x$ for some $\epsilon>0$ (epsilon characterization of the supremum)

This is not the epsilon characterization of the supremum. The characterization says that:

$s$ is a supremum of $A$ if, for every $\epsilon > 0$, there exists some $x\in A$ such that $s-\epsilon < a$.

This is completely different from what you wrote. Most importantly, the definition of supremum says something is true for every epsilon, not "for some $\epsilon > 0$ which is what you wrote.
Problem 2:
You wrote "$\forall x \in b$", which makes no sense. $b$ is not a set, so $x\in b$ is nonsensical.
A: The definition of an upper bound $y$ being a supremum of the set $A$ is that for all $\epsilon >0$, there exists $x \in A$ such that $y < x+\epsilon$. Your quantifiers are not correct.
The converse to this statement then (that is if $y$ is not a supremum) is that there exists some $\epsilon >0$ such that for all $x \in A$, $x+\epsilon \leq y$.
So a proof by contradiction would start:
"Suppose that $b$ is an upper bound of $[a,b)$ but $b$ is not the supremum of $[a,b)$. Then there exists an $\epsilon >0$ such that for all $x \in [a,b)$, $x+\epsilon \leq b$. Then, ..."
A: You have the right idea, but your formulation of the epsilon chracterization of the supremum isn't correct. It should be
$$\forall \epsilon > 0\quad \exists x\in [a,b) : x+\epsilon > b\quad (+)$$
The negation of that would be
$$\exists \epsilon > 0\quad \forall x\in [a,b): x+\epsilon \leq b\quad (*)$$
By choosing $x$ in a suitable way, you can now show that the inequality $(*)$ can't hold for every $x\in [a,b)$. For example, define $x$ by $x:=\text{max}\{a, b-\frac{\epsilon}{2}\}\in [a,b)$. It follows that
$$x+\epsilon \geq b-\frac{\epsilon}{2} + \epsilon = b+\frac{\epsilon}{2} >b,$$
which is a contradiction.
Also, the negation of $x+\epsilon > b\geq x$ isn't $x+\epsilon \leq b<x$.
