Is the set of real numbers the largest possible totally ordered set?
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.