Prove that if $A$ has a maximum, then the maximum must be the supremum 
Prove that given $A\subseteq\mathbb{R}$, if has a maximum $a_0$, then $a_0$ must be the supremum.

In order to prove this, I have written the following:
Let $A\subseteq \mathbb{R}$, nonempty set because $a_{0}$ $\in A$ for hypotesis. Then, since $a_{0}$ $\in A$ is the largest element of the set $A$   (i.e. for each element $A$ of A we have that $ a\leq a_{0}$) $A$ is bounded above. Then, A has a supremum. Let $s$ be the least upper bound of $A,$ so we have that for each element $a$ of $A$, $a\leq s$, for being $s$ an upper bound. In particular, $a_0\leq s$, which implies that $s=a_0$. 
Then, we have that $\sup(A)=\max(A)=a_0$, which is exactly what we wanted to prove.
Observation: this proof is true if $s \in A$ 
I'm not sure if my proof is correct. I will be thankful if someone tells me.
 A: No you're assuming that $A$ includes it's supremum. You want to do something like the following:
Let $A \subseteq \mathbb R$ be a set such that $a_0 \in A$ is the largest value in $A$. Let $x \not\in A$, then we find that either $a_0 > x$ and thus $x$ is not an upper bound or $a_0 < x$ in which case $x$ is an upper bound but $a_0$ is an upper bound that is less then it, then for any number $x\not\in A$, $x\not=\sup A$, thus $\sup A \in A$ and then you're proof will show that the supremum must be $a_0$.
The issue with your proof was that you didn't show that the supremum was contained within $A$, you just kind of assumed it.
A: To be frank this result is typical of many theorems in analysis which are as trivial as they can be and your approach is unnecessarily complicated and based on the unfounded assumption that things in analysis are in general non-trivial.
The result is an immediate consequence of the definitions of maximum and supremum. Let $a_{0}$ be the maximum of set $A$. Then by definition of maximum we have $a\leq a_{0}$ for all $a\in A$. Moreover if we take any number $b<a_{0}$ then it's just repeating the same thing that we have $a_{0}\in A$ and $a_{0}>b$. Thus $a_{0}$ is the supremum of $A$.
There is no need to show that $A$ is bounded above and invoke deep completeness property to justify the existence of $\sup A$. You can see that the above argument (and therefore the result in question) holds in any ordered field (eg $\mathbb{Q}$).
It is rather surprising that other answers here also make use of the completeness property unnecessarily. 
A: You have not be given the fact that $A$ is bounded from above.  It is not empty because it contains $a_0$.  Since $a_0$ is its maximum, then $A$ is bounded from above by $a_0+1$.  Hence it is not empty and it is bounded from above. 
 Since it is a subset of $\mathbb{R}$, it has a supremum.  
Let $b$ be the supremum of $A$.  Since $a_0\in A$, $a_0 \le b$.  Now you need to show that $a_0 \ge b$.
Let $\epsilon > 0$ be given.  Since $b$ is the supremum of $A$, there is an $x \in A$ such that $b - \epsilon \le x$.  But then
$$
b - \epsilon \le x \le a_0
$$
since $a_0$ is the maximum of $A$.  Since $\epsilon$ was arbitrary, we can conclude that
$$
b \le a_0
$$ 
Hence $a_0=b$.
A: Let $b := sup A$, since $a_{0} \in A$ (is maximum), then $b \geq a_0$. Now we need to prove that $b \leq a_0$, for this suposse the contrary, this is $b > a_0$: As $a_0 \in A$ we have that $a_0 \geq a$ for all $a \in A$, then $a_0$ is upper bound of $A$, thus we have:


*

*$b = sup A$

*$b \geq a_0$

*$a_0$ is upper bound


So contradiction! xd. Therefore $b \leq a_0$, finally $b = a_0$.
