Is the set of totally ordered sets totally ordered?

The natural numbers, $\omega$ and their inverse order (i.e. the negative integers) are incomparable.

But this is not even a partial order. The rational numbers embed into $[0,1]\cap\Bbb Q$ and vice versa, so it's not even antisymmetric.

The answer is no, for example there is no order preserving injection $\omega_1\to\Bbb R$ (every ordinal embedding into the reals is countable) as well as no order preserving injection $\Bbb R\to\omega_1$ (since $\Bbb R$ has infinite descending chains, while $\omega_1$ doesn't)