Proof of the greatest integer theorem: for every real number $x$ there exists a unique greatest integer less than or equal to $x$ To define the function $f(x)=|[x]|$ where $|[x]|$ is the greatest integer that is less or equal to $x$, we need to prove that indeed such an integer exists. In other words,

$$\forall x\in \mathbb{R} \;\;\exists !\, n\in\mathbb{Z}\;\;:\;\;n\leq x<n+1.$$

My first attempt is by defining the set $A=\{y\in\mathbb{Z}| y\leq x\}$. Since $\mathbb{N}$ is not bounded above then $\exists n\in \mathbb{N}(n>-x)$. Then $\exists n'\in \mathbb{Z}(n'<x)$, namely $n'=-n$. Therefore $A\neq\emptyset$. Also $A$ is bounded above by $x$. Then $\sup(A)$ exists. 
**Here my problem is how to prove that $\sup(A)\in A$. 
Also my question is that I'm not sure if this is enough because $sup(A)$ is unique or do I still need to prove uniqueness.  

EDIT:
After all the hints in the answers, chich I appreciate so much, I have arrived to this attempts of solution, please feel free to comment and help me saying if something is wrong: 
First approach: considering tree cases ($x<0, x=0, x>0$). 
1) If $x=0$ then I can think of $n=0$. 
2) If $x>0$ then I can think of the set $B=\{y\in \mathbb{N}|x\leq y\}$. By using the well-ordering principle then $\min(B)-1<x\leq \min(B)$ and then if $x=\min(B) $ I can make $n=\min(B)$; if  $\min(B)-1<x<\min(B)$ then $n=\min(B)-1$. 
3) if $x<0$ then $-x>0$ and therefore $\exists m \in \mathbb{N}(m\leq -x < m+1)$. Then $\exists n'\in \mathbb{Z}(-n'<x\leq-n'-1)$, namely $n'=m+1$. Again, if $x=-n'-1$ then I can make $n=-n'-1$. If $-n'<x\leq-n'-1$ then $n=-n'$.
Second approach. Since $\mathbb{N}$ is not bounded above then let $x_{0}\in \mathbb{N}$ be such that $x_{0}>\sup(A)$. Let's consider the set $B=\{x_{0}-y: y\in A\}$. Since $B\subseteq \mathbb{N}$ then, by the well ordering principle, $\min(B)$ exists. Moreover $\exists y_{0}\in A\forall y\in A(\min(B)=x_{0}-y_{0}\leq x_{0}-y)$. This means that $\exists y_{0}\in A\forall y\in A (y\leq y_{0})$. Therefore $\max(A)=y_{0}$. In this case I can make $n=\max(A)$.   
 A: Hint: Rather than dealing with $A=\{y\in\mathbb{Z}\mid y<x\}$, try dealing with $B=\{y\in\mathbb{Z}\mid y\leq x\}$. 
Why?  In general, it is not true that $\sup A\in A$; it is true whenever $x\notin\mathbb{Z}$, but it is false when $x\in\mathbb{Z}$.
Now, think about it this way: let $S:=\sup B$. Since $S$ is the supremum, it must be the case that there is $n\in B$ such that $S-1<n\leq S$.
Can you show that $S\leq x$? If so, you'll have that $n\leq x$. For the other side, note that $S-1<n$ implies $S<n+1$, which implies that $n+1\notin B$; hence $n+1>x$.
A: First, do you know what real numbers are? In general, you can discuss real numbers without knowing exactly what they are, but I think that when trying to justify some basic properties it is good to be familiar with their definition. There are some equivalent constructions, and if you don't know them I encourage you to check them out.
As for your question, I think that a good approach is to prove the theorem for rational numbers and then deduce it to real numbers since every real number is a limit of rational numbers.
A: $n\le z<n+1$ is not the definition of greatest integer less than or equal to $z$, it's a consequence.
You let $A=\{\,y\in \mathbb Z\mid y\le z\,\}$.
You use the Archimedean property (twice) to show $\sup A$ exists.
The supremum of a subset of $\mathbb R$ is always unique.
To show that $\sup A\in A$ it would be sufficient to show that $A$ is closed, which is the case because $A\subseteq \mathbb Z$.

An alternative method (for brevity here described only for $z>0$) is to note that $\mathbb N\setminus(-\infty,z]$ is nonempty by the Archimedean property, has a smallest element $n$ (by well-order of $\mathbb N$) and that therefore $n-1\le z<n$. From this it follows  $n-1$ is the largest integer $\le z$.
A: You don't need the $\sup$ for establishing this. 
Any construction or description of ${\mathbb R}$ establishes the following facts: $({\mathbb Q},<)\subset({\mathbb R},<)$, and given a real number $\xi$ there are two rational numbers $a$ and $b$ with $a<\xi<b$. Making $a$ a little smaller and $b$ a little bigger we may even assume $a$, $b\in{\mathbb Z}$. The set $\{k\in{\mathbb Z}\>|\>a\leq k\leq b\}$ is finite, and as $\xi<b$  the set $S:=\{k\in{\mathbb Z}\>|\>a\leq k\leq \xi\}$ is finite as well. Therefore $S$ has a maximal element, which we denote by $\lfloor\xi\rfloor$.
