I may be missing something, but it seems to me that the exact same statement holds for all infinite total orders.

Indeed, if $\mathbb Q$ does not embed in $\Omega$, then $\Omega$ is scattered, hence it can be constructed from the singleton order by repeatedly using well-ordered and inverse well-ordered sums. Then we can prove by induction on this construction that once the order is infinite, it satisfies the first clause: if $\Omega=\sum_{\alpha<\gamma}\Omega_\alpha$ is infinite, then either some $\Omega_\alpha$ is already infinite, or not. In the first case, $\Omega_\alpha\simeq\Omega_\alpha\smallsetminus\{x\}$ for some $x\in\Omega_\alpha$ by the induction hypothesis, hence $\Omega\simeq\Omega\smallsetminus\{x\}$ as well. In the second case, $\Omega$ itself is an infinite well order, hence it is isomorphic to itself minus its first element. The argument for inverse well ordered sums is completely analogous.

As YCor pointed out, this argument actually shows a bit more: if $\Omega$ is any infinite total order, then

BTW, I had a look at the Dushnik and Miller paper (Concerning similarity transformations of linearly ordered sets, Bull. AMS 46 (1940), no. 4, pp. 322–326) to see where did countability come from. It turns out they do not state the result formulated in the question at all. Rather, it only implicitly follows from their proof of Theorem 1, which actually states that *every countable infinite total order is isomorphic to a proper subset of itself*. This statement does *not* hold for uncountable orders (the paper includes a counterexample).