We just don't have that many examples of models where the reals have a Hamel basis but no well-ordering. Mind you, we know of a number of those, but that's a small number indeed.
One way to solve this is to note that if $M$ is model of $\sf ZF$ in which there is a Hamel basis, and $M\subseteq N$ is such that $\Bbb R^M=\Bbb R^N$, then $N$ has a Hamel basis.
So a good strategy is to start with a model where a Hamel basis exists and add via forcing (or symmetric extensions) a countable family of countable sets whose union is uncountable. Note that the above means that we are not adding reals, so the countable sets need to already exist in the model, adding the sequence of these sets is not a problem, since the union is uncountable, this additional sequence will not code any new reals.
The natural knee-jerk reaction is to try and use the Cohen model, which famously has a Hamel basis (although that is quite a recent advance), but in the Cohen model the union of a well-orderable family of well-orderable sets is well-orderable, so that will not work. It might be possible to use the Cohen model as our ground and add the aforementioned sequence.
Of course, another alternative is to simply prove directly that some model where there is a countable family of countable sets of reals whose union is uncountable does have a Hamel basis.