Proving rigorously the supremum of a set Suppose $\emptyset \neq A \subset \mathbb{R} $.  Let $A = [\,0,2).\,\,$   Prove that $\sup A = 2$
This is my attempt:

$A$ is the half open interval $[\,0,2)$ and so all the $x_i \in A$ look like $0 \leq x_i < 2$ so clearly $2$ is an upper bound.
To show it is the ${\it least}$ upper bound, suppose that $2 \neq \sup A$, that is there exists a number $M < 2$ for some real $M$ qualifying as $\sup A$.  Certainly this $M \in [0,2) $ so $ M > 0 \Rightarrow 2 -M > 0$.
By the Archimedean Principle, for all real numbers $r > 0\,\, \exists\,\, n \in \mathbb{N}$ such that $0 < \frac{1}{n} < r $. By the Approximation Property of Suprema, there exists $a \in [0,2)$ such that $\sup A - \epsilon < a \leq \sup A$, where $\epsilon > 0$.
Suppose $\sup A = M < 2$.  Then the above gives $M - \epsilon < a < 2\,\,\,\,\forall \epsilon > 0$.  Also, by Archimedean, we have $0 < \frac{1}{n} < 2-M$, so choose $\epsilon = 2-M$.  Then $M - (2-M) < a < 2 \,\Rightarrow 2(M-1) < a < 2$
We can assume $M - 1>0$ and so $2(M-1) > 2$  This results in a contradiction in the previous inequality.  Hence $M < 2$ cannot be the supremum.

I realise there is probably a simpler way, but is what I have written all good?
 A: 
Let $a\lt 2$  is $\sup A$. 
Therefore,  $a=2-b$ , where $b\gt 0$.
We can get some $n\in \mathbb{N} \,|\, 0 \lt \frac1n \lt b$
So, $2-b\lt2-(\frac1n)\lt2$. 
$\exists c\in Q\cap A\,|\,2\gt c\gt 2-(\frac1n)\gt 0$.
So,$\,2\gt c \gt 2-(\frac1n)\gt (2-b)=a$.
$0\lt c \lt 2 \implies c\in A$, and also $c\gt a$
i.e. $a$ is not even an upper bound of $A$. 
$\implies$ the set of upper bounds of $A$ is $\{x : x\in \mathbb{R}$ and $x\ge 2\}=S$ and the least member of $S$ is $2$.
So, $2$ is the least upper bound of $A$.
A: I think there's a way more simple and intuitive proof. 
First, as you observed, it is obvious that 2 is an upper bound. Now, to prove it is the supremum. Assume that $M$ is the supremum and $M<2$. 
Of course, $M>2$ is trivially impossible since $2$ is an upper bound as well and thus any $M$ bigger than $2$ cannot be a supremum.
Now, let $x=\frac{2-M}{2}+M$.
As you can obviously see $M<x$, so all that is left is to show that $x<2$.
But now, assume it is not, that is $x\geq2$. 
Then, $\frac{2-M}{2}+M\geq2$. Multiplying both sides by $2$, we get
$2-M+2M\geq4$, that is $2+M\geq4$. But $M<2$, so $2+M<4$. Thus, by reductio ad absurdum, $x<2$. This shows that $M<2$ cannot be the supremum. 
QED.
Now, as for the simplicity of this proof, I have written a lot for clarity and in case you are a beginner on this subject. This can be summarized in $2$ lines, but this is for clarity. I hope this helps, and you must soon learn to find the shortest and more intuitive way. Good luck.
