Let us say that two sets $A$ and $B$ are comparable if there is an injection from $A$ to $B$ or there is an injection from $B$ to $A$. Obviously, in a model of ${\rm ZFC}$ any two sets are comparable by comparing their cardinalities. But this is not necessarily the case in a model of ${\rm ZF}$. For example in Cohen's classical symmetric submodel showing the independence of the Axiom of Choice from ${\rm ZF}$, there is a set of reals which is incomparable with $\omega$. 


Is there a known construction of a model of ${\rm ZF}$ in which the reals are not well-ordered, but any two sets of reals are comparable? 



A model of ZF without a well-ordering of the reals in which any two sets of reals are comparable
Yes. The perfect set property will ensure every set of reals is countable or has size continuum.



> 
> *Solovay, R.M.*, [**A model of set-theory in which every set of reals is Lebesgue measurable**](http://dx.doi.org/10.2307/1970696), Ann. Math. (2) 92, 1-56 (1970). [ZBL0207.00905](https://zbmath.org/?q=an:0207.00905).
> 
> 
> 


In the paper Solovay proved that from the consistency of an inaccessible cardinal it is consistent that every set of reals is countable or has the perfect set property. Of course, there you need the inaccessible cardinal if you want Dependent Choice. If you're willing to forego that, you can forego the inaccessible cardinal, as shown by Truss.



> 
> *Truss, John*, [**Models of set theory containing many perfect sets**](http://dx.doi.org/10.1016/0003-4843(74)90015-1), Ann. Math. Logic 7, 197-219 (1974). [ZBL0302.02024](https://zbmath.org/?q=an:0302.02024).
> 
> 
> 


There he shows that you can always push this up in the sense that you can have that every set of reals is well-orderable or has a perfect subset and you can pretty much have no limit on what "well-orderable" means in that case. (Although I believe Solovay also does that at the end of his paper, but I do not recall at the moment.)


