Let's say that a linear order is $\omega_1$-short if it has no uncountable well-ordered or reverse well-ordered subset. For instance, linear orders satisfying Suslin's condition in his hypothesis are $\omega_1$-short. There are $\omega_1$-short linear orders which do not satisfy the countable chain condition.
Is there a $\omega_1$-short linear order in which every $\omega_1$-short linear order embeds?
Any $\omega_1$-short linear order embeds in the complete lexicographically ordered binary tree $2^{<\omega_1}$ (which can be ssen as a subfield of the ordered Field of surreals), but $2^{<\omega_1}$ is not $\omega_1$-short. I do not think that any of the $2^{<\alpha},\alpha < \omega_1$ qualify, even though they are $\omega_1$-short, but I don't have a proof. I actually don't know yet if $2^{<\alpha+1}$ can embed in $2^{< \alpha}$ for some values of $\alpha < \omega_1$.