Why does$A\subset \mathbb{Z}$ have a maximum value if it bounded from above? Why does it happen that, if $A\subset \mathbb{Z}$ is bounded from above, it has a maximum value? How can it be explained?
 A: It is more convenient to show that if $A\subset \mathbb{Z}$ is bounded from below (rather than above) then there is a minimum.  Namely, if $A$ is bounded from below then there is an integer $n$ such that $A+n$ is bounded from below by $0$.  Thus we can assume wlog that $A$ is a collection of natural numbers.  But the natural numbers are well-ordered in the sense that any subset has a least element.
A: I would go as follows:
Bounded means that there exists $N\in\Bbb Z$ such that $A\subseteq\{n\in\Bbb Z\,\mid\,n<N\}$.
For eack $k\in\Bbb N$, let 
$$I_k=\{n\in\Bbb Z\mid N-k\leq n\leq N\}$$
and set
$$
\Sigma=\{k\mid I_k\cap A\neq\emptyset\}.
$$
Assuming $A\neq\emptyset$, the set $\Sigma$ is not empty because $A\subseteq\bigcup_kI_k$ and so it has a minimum $m$. But then $N-m=\max A$.
A: Assume $A\ne\emptyset$ and that there is an $M\in{\mathbb Z}$ with $x\leq M$ for all $x\in A$. Put
$$B:=\{y\in{\mathbb N}_{\geq0}\>|\>M-y\in A\}\ .$$
Then $B\ne\emptyset$; therefore $B$ has a minimal element $y_*$. I claim that $x_*:=M-y_*\in A$ is the maximal element of $A$. Proof: For any $x\in A$ one has $y:=M-x\in B$ and therefore  $$x=M-y\leq M-y_*= x_*\ .$$
