Is the set of real numbers the largest possible totally ordered set?
Because I find any totally ordered set can be "lined up" in a straight line, I'm guessing that the set of all the real numbers is the biggest totally ordered set possible. In the sense that any other totally ordered set is isomorphic (order-preserving) to a subset of real numbers. Is that right?
Elementary Construction
Without going into too much set theory, it is possible to produce an explicit example of a totally ordered set that cannot be embedded into $ (\mathbb{R},\leq_{\mathbb{R}}) $ in an order-preserving manner.
Define a total ordering $ \preceq $ on $ \mathbb{R}^{2} $ as follows: $$ \forall a,b,c,d \in \mathbb{R}: \quad (a,b) \prec (c,d) ~ \stackrel{\text{def}}{\iff} ~ (a <_{\mathbb{R}} c) ~~ \text{or} ~~ (a = c ~~ \text{and} ~~ b <_{\mathbb{R}} d). $$ Observe that $ \{ \{ r \} \times \mathbb{R} ~|~ r \in \mathbb{R} \} $ is an uncountable collection of pairwise-disjoint non-degenerate $ \preceq $-intervals of $ \mathbb{R}^{2} $. Hence, $ (\mathbb{R}^{2},\preceq) $ cannot be embedded into $ (\mathbb{R},\leq_{\mathbb{R}}) $ in an order-preserving manner; if this were not the case, then $ \mathbb{R} $ would contain uncountably many pairwise-disjoint non-degenerate intervals, so by picking a rational number from each of these intervals, we would end up with uncountably many rational numbers ⎯ contradiction.
Model-Theoretic Construction
There is a neat way to prove the existence of totally ordered sets with arbitrarily large cardinality. It relies on the Compactness Theorem in logic and model theory. Let $ \kappa $ be a cardinal and $ \mathcal{L} $ the first-order language consisting of
a set $ \mathcal{C} $ of $ \kappa $-many constant symbols and
a single binary relation symbol $ \leq $.
Next, define $ T $ to be the first-order $ \mathcal{L} $-theory consisting of:
For distinct $ c,d \in \mathcal{C} $, the sentence $ \neg(c = d) $.
The Axiom of Reflexivity: $ (\forall x)(x \leq x) $.
The Axiom of Antisymmetry: $ (\forall x)(\forall y)(((x \leq y) \land (y \leq x)) \Longrightarrow (x = y)) $.
The Axiom of Transitivity: $ (\forall x)(\forall y)(\forall z)(((x \leq y) \land (y \leq z)) \Longrightarrow (x \leq z)) $.
The Total-Ordering Axiom: $ (\forall x)(\forall y)((x \leq y) \lor (y \leq x)) $.
Although $ T $ has infinitely many sentences (by (1)), it is easy to see that $ (\mathbb{N},\leq_{\mathbb{N}}) $ is a model of any finite number of these sentences. Therefore, by the Compactness Theorem, $ T $ has a model $ (\mathcal{M},\leq^{\mathcal{M}}) $. Clearly, if we choose $ \kappa > {\frak{c}} $, then we cannot embed $ (\mathcal{M},\leq^{\mathcal{M}}) $ into $ (\mathbb{R},\leq_{\mathbb{R}}) $ in an order-preserving manner.
Set-Theoretic Construction
The model-theoretic approach above suffers from a mild drawback, in the sense that the Compactness Theorem is equivalent to the Boolean Prime Ideal Theorem, which is a weak form of the Axiom of Choice. If one wishes to work strictly within the framework of the Zermelo-Fraenkel axioms, then one can use Hartogs’ result that for any set $ X $, it is possible to find an ordinal $ \alpha $ that cannot be mapped injectively into $ X $. In particular, there exists an ordinal $ \alpha $ that cannot be mapped injectively into $ \mathbb{R} $, much less embedded into $ \mathbb{R} $ in an order-preserving manner.
No, absolutely not. Assuming the axiom of choice, let $\kappa$ be any cardinal greater than $2^\omega=|\Bbb R|$; then $\kappa$ in its natural order is a linear order of greater cardinality than $\Bbb R$. If the axiom of choice fails in such a way that $\Bbb R$ cannot be well-ordered, there is still a well-ordered set that cannot be embedded in $\Bbb R$, the Hartogs ordinal of $\Bbb R$.
The real line is the largest order separable linearly ordered set. That is, if $(L,\preceq)$ is a linearly ordered set such that a countable set $C\subseteq L$ exists satisfying that for all $x,y\in L$ with $x\prec y$ there is $c\in C$ with $x\preceq c\preceq y$, then there exists a function $f:L\to\mathbb{R}$ such that $f(x)<f(y)$ iff $x<y$ for all $x$ and $y$ in $L$. The existence of such a set $C$ is also necessary. A proof of these facts is given here.
Here is yet another counterexample without order separability: Let $(W,\preceq)$ be an uncountable well-ordered set. Let $W'$ be $W$ without the maximum of $W$ if such a maximum exists. For each $w\in W'$, let $S(w)$ be its immediate sucessor given by $S(w)=\min\{w'\in W:w'\succ w\}$. It is easily seen that $x\prec y$ implies $x\prec S(x)\preceq y\prec S(y)$. So if a strictly increasing function $f:W\to\mathbb{R}$ would exist, the uncountable family of intervals $\{(f(w),f(S(w))):w\in W'\}$ would be disjoint, which cannot be for the same reason pointed out by Haskel Curry in his answer.
EDIT: this may be order isomorphic to the reals, but not as a field, as it is not Archimedean.
As an elementary example, the rational functions $\mathbb R(x)$ are an ordered field. The quotient $p(x) / q(x),$ with leading terms $p_m x^m$ and $q_n x^n,$ is called positive if $p_m / q_n$ is positive. Comparison of two rational functions is given by taking the difference and checking the sign as above. So, for example, $1/x$ is positive but is smaller than any positive real number, as with positive $A$ we find $$ A - \frac{1}{x} = \frac{Ax-1}{x}. $$