If $\mathcal{L}$ ={$\leq$}. A linear order $\mathcal{A}$= (A , $\leq$) is discrete if and only if
I- $\forall$ $a\in A$ there is some $a'\in A$ such that $ a< a'$ then there is a minimal $b\in A$ which $ a< b$.
II- $\forall$ $a\in A$ there is some $a'\in A$ such that $ a' < a $ then there is a maximal $b\in A$ which $ b < a$.
Intuitively, we can identify 4 classes of (infinite) Discrete Linear Orders. Namely
1-The class of discrete linear orders with a minimal element and without a maximal element. As an example ( $\mathbb{N},\leq$).
2-The class of discrete linear orders with a maximal element and without a minimal element. As an example ( $\mathbb{N}$, something like $\geq$).
3-The class of discrete linear orders with both a maximal element and a minimal element. As an example, one can say the union of two discrete linear orders (A, $\leq$) $\cup$ (B, $\leq$). [ such that the first one is like case 1, and the second one is like case 2. Plus, for all $a\in A$ and $b\in B$, $a< b$ ].
4-The class of discrete linear order without endpoints. As an example ( $\mathbb{Z},\leq$).
How do I write a proof for this intuition? And then show any other arbitrary discrete linear order would necessarily fall into one of these classes.
$\it {Edit 1: }$ I'm not sure about maximal and minimal words, maybe last and least is better. [ Or top and bottom which I saw in a version ].
$\it {Edit 2: }$ Regarding to Alex Kruckman's comments, I modify my question :
How can one prove every infinite countable discrete linear order necessarily must be isomorphic to one of these four structures?
Best Answer
You suggest dividing the infinite discrete linear orders into four classes:
As pointed out by spaceisdarkgreen in the comments, it is a tautology that every discrete linear order falls into one of these four classes: every linear order either has a maximal element or doesn't, and every linear order either has a minimal element or doesn't. This is a complete answer to the original question.
It is not a tautology, but also not hard to prove, that these four classes correspond to the four completions of the theory of infinite discrete linear orders. That is, if you take the theory $T$ which says:
then adding any one of the axioms defining classes (1) - (4) gives a complete theory. Let's call these theories $T_1$, $T_2$, $T_3$, $T_4$.
You also give a canonical example of a model of each of these theories:
It turns out that each of these models is the prime model of its theory (it admits an elementary embedding into any other model).
From completeness of $T_1$ through $T_4$, and the (tautological) fact that every infinite discrete linear order is in one of your classes 1 through 4, it follows that every infinite discrete linear order is elementarily equivalent to one of the four structures $\mathbb{N}$, $\mathbb{N}^*$, $\mathbb{N}+\mathbb{N}^*$, or $\mathbb{Z}$, depending on whether it is a model of $T_1$, $T_2$, $T_3$, or $T_4$.
In the follow-up question, you ask:
You cannot prove this, because it is false. I gave you counterexamples in the comments. Another way to put it is that none of the theories $T_i$ is countably categorical - this is the contrast with the case of dense linear orders. If you take the theory of dense linear orders and specify whether there is a maximal and/or a minimal element, then you get a complete and countably categorical theory.
Nevertheless, we can obtain a reasonable understanding of the structure of all models of the theories $T_i$. When $L$ and $L'$ are linear orders, by $L+L'$ I mean the disjoint union of $L$ and $L'$, with all the element of $L'$ coming after the elements of $L$. By $L\times L'$, I mean the cartesian product of $L$ and $L'$, with the lexicographic order.
Note that your canonical examples are given by taking $L$ to be empty in cases (1) - (3), and by taking $L$ to be the one-point linear order in case (4).
Note also that whenever $L$ is countable, the resulting model is countable. Since there are continuum-many countable linear orders up to isomorphism, it follows that each of the theories $T_i$ has countinuum-many countable models up to isomorphism. This is as far as possible from your claim that every countably infinite discrete linear order is isomorphic to one of your canonical structures.
Here's a plan for how you can prove the classification: Let $M$ be an infinite discrete linear order. Define a relation $a\sim b$ if and only if $b$ is the successor of $a$. Let $E$ be the reflexive symmetric and transitive closure of $\sim$. Prove that $E$ is an equivalence relation with convex classes. Then the linear order on $M$ naturally orders the $E$-classes as some linear order $L$. Show that the minimal class (if there is one) is ordered like $\mathbb{N}$ and the maximal class (if there is one) is ordered like $\mathbb{N}^*$, and every other class is ordered like $\mathbb{Z}$.