The Order of sets like $\mathbb Q$ and Justification of the Number Line Rudin, in Principles of Mathematical Analysis, defines an ordered set $S$ as a set with a relation such that 

(i)  If $x\in S$ the one and only one of the statements $$x<y,\,\,\,\,x=y\,\,\,\,\,y<x$$
  is true.
(ii) For $x,y,z\in S$, if $x<y,$ and $y<z$, then $x<z$.

He also gives the axioms for an ordered field (field axioms, including axioms for (+) and ($\bullet$) for inequalities).
My question is: Are the above two properties of (<) all we need for a set to geometrically represent it as points on a line such that for any point $x$ if a point $y$ is to the right of $x$ then $x<y$? Are all ordered sets ($S$,<) isomorphic if say $S$ is countable? Can all sets that satisfy $(i)-(ii)$ be put into a line similar to $\mathbb Q$?
If not, what other order properties are needed to justify a linear graph of the the rational or real numbers? 
:Edit: What field of study would the justification of the number line be? Topology? Order Theory?
 A: You're touching on the field of "order-theory" (which has connections to model theory, topology  and set theory, mostly). The axioms describe a (strict) linear order, and the standard examples are $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$ which almost anyone doing mathematics encounters somehow. All these have a standard linear order associated with them.
In general we consider two linearly ordered sets $(X,<_X)$ and $(Y,<_Y)$ to be order-isomorphic when there exists a bijection $f: X \to Y$ such that $$\forall x,x' \in X: x <_X  x' \iff f(x) <_Y f(x')$$
and such sets are considered to have "the same order structure". Abstractly we have a partition into linearly ordered sets into "equivalence class" of this with the same order, and such an equivalence class is called an "order type". We can talk about the order type of $\mathbb{N}$, as all linear orders that are order isomorphic to $\mathbb{N}$ (intuitively: start with a minimal element, and add a unique successor for every element, and don't stop (or unit we first have an infinite number of elements). A boring example: the even numbers in their inherited order are the same order type as $\mathbb{N}$, and some thought will show that the same holds for all infinite subsets of $\mathbb{N}$. For every finite number $n$ there is a unique order type of linear orders of type $n$, but this is definitely not the case for countable sets.
Some order properties for some countable orders that can serve to show they're not order-isomorphic:


*

*$\mathbb{N}$ has a minimum, $\mathbb{Z}$, $\mathbb{Q}$ do not.

*$\mathbb{Z}$ has the property that every point has two neighbours (a right neighbour of $x$ is an $x^+$ such that $x < x^+$ but not $x'$ exists with $x < x' < x^+$; a left neighbour is the same but smaller), while $\mathbb{Q}$ does not have that property  and in $\mathbb{N}$, $0$ has no left neighbour.

*$\mathbb{Q}$ is order dense ($\forall x,y \in X: (x < y) \to (\exists z \in X: x < z < y)$ but the orders $\mathbb{N}$ and $\mathbb{Z}$ are not (see previous property).
It turns out that the order type of $\mathbb{Q}$ is special: If $(X,<)$ is a countable linearly ordered set, and $X$ is order dense and has no maximum or minimum, then $X$ is order isomorphic to $\mathbb{Q}$, and moreover any countable linearly ordered set is order-ismorphic to some subset of $\mathbb{Q}$ (in the inherited order from the rationals). So all countable can certainly be embedded into a number line (namely $\Bbb Q$), but for larger cardinals this need not be true, I believe. For a little more on order types see its Wikipedia page, e.g. 
The well-ordered (meaning: every non-empty subset has a minimum) linear orders get a special study in set theory. They are uniquely represented by so-called ordinal numbers. They form the basis for cardinal numbers.
And $\Bbb Q$ can be so-called "order-completed" (there are still gaps in it: sets with upper bound but with no least upper bound) and it turns out the order type of the order-completion of $\Bbb Q$ is exactly $\Bbb R$ and in analysis this order compeleteness is used all over the place. 
A standard work on linear order theory see the book "Linear Orderings", by Rosenstein. You can define addition and multiplication of order types e.g. Any linearly ordered set has a natural topology too (in fact for all four examples in the first paragraph, their natural topologies are in fact the order topology), that I recently talked about here, if you're interested. 
A: I’m not exactly sure what you mean by “isomorphic,” but I assume that $(S_1, <_1)$ and $(S_2,<_2)$ are said to be “isomorphic” if there exists a bijection $f:S_1\mapsto S_2$ such that $x_1 <_1 y_1\iff f(x_1) <_2 f(x_2)$ for all $x_1,y_1\in S_1$ and $x_2,y_2\in S_2$. I also assume that these orderings $<_1, <_2$ satisfy $x<y\iff y>x$.
I under this definition, not all countable sets $S$ are isomorphic. For example, there cannot exist any such order-preserving bijection from $\mathbb Z$ to $\mathbb Q$, or for that matter, from $\mathbb N$ to $\mathbb Z$.
We can intuitively prove that no such “isomorphism” $f:\mathbb N\mapsto\mathbb Z$ exists because for all $n$ in $\mathbb Z$ under the natural ordering, there exists $n’$ such that $n’ < n$, while this is not the case for $\mathbb N$. 
We can also intuitively prove that no “isomorphism” from $\mathbb Z$ to $\mathbb Q$ exists because for all $p,q\in\mathbb Q$ satisfying $p<q$ under the natural ordering, there exists $r\in\mathbb Q$ such that $p<r<q$, but this property does not hold for $\mathbb Z$.
