If a set is equipped with total order (reflexive, transitive, antisymmetric, strongly connected), does this necessarily mean there exists an injective mapping from that set to the reals? Alternatively, if a set has larger cardinality than the reals can it not be well-ordered?
I'm struggling to think of a counter example. Most of the canonical examples of sets with larger cardinality than the reals (all functions, power set of reals) cannot be equipped with total order.
If we assume the axiom of choice then - contra your claim "Most of the canonical examples of sets with larger cardinality than the reals (all functions, power set of reals) cannot be equipped with total order" - every set can be linearly (indeed well-) ordered.
Even in the absence of choice, though, we can prove that the answer to your question is negative via Hartogs' theorem, which says that there is an ordinal $\alpha$ with no injection to $\mathbb{R}$. (More generally, for any set $X$ there is an ordinal not injecting into $X$.) The proof is quite simple: let $\mathcal{A}$ be the set of well-orderings of subsets of $\mathbb{R}$, up to order-isomorphism, and note that $\mathcal{A}$ itself is well-ordered by $[R]\le[S]$ iff $R$ is isomorphic to an initial segment of that corresponding to $S$. The unique ordinal $\mathcal{A}$, well-ordered this way, is isomorphic to will not embed into $\mathbb{R}$. Note that this proof does use Powerset and Separation, but does not use Choice.
(And if we replace "well-order" with "pre-well-order" in our definition of $\mathcal{A}$ we get an even stronger result: there is an ordinal onto which $\mathbb{R}$ does not surject! In the absence of choice this is not the same as simply lacking an injection into $\mathbb{R}$, and indeed the two ordinals so produced can be wildly different: if the axiom of determinacy holds, for instance, then $\omega_1$ does not inject into $\mathbb{R}$ but the smallest ordinal $\mathbb{R}$ doesn't surject onto - called "$\Theta$" - is much bigger than $\omega_1$.)