$\neg \textsf{AC}+ \neg\textsf{CH}$

Question

1. Is there some interesting\surprising results that have only been proven by assuming $\neg \textsf{AC}$ and $\neg\textsf{CH}$ ?
2. Is there some interesting\surprising results implying both $\neg \textsf{AC}$ and $\neg\textsf{CH}$ ?

Edit :

My definition of $\textsf{CH}$ in $\textsf{ZF}$ is $\aleph_0<\mathfrak m\rightarrow 2^{\aleph_0}\leq\mathfrak m$

Answer 1

If you consider the continuum hypothesis to be $\aleph_0<\mathfrak m\rightarrow 2^{\aleph_0}\leq\mathfrak m$, then one possible result worth mentioning is:

There exists an infinite Dedekind-finite set of reals which is Borel.

If $A$ is such set, then $A\cup\Bbb N$ is a witness for the failure of $\sf CH$ and $\sf AC$. What is perhaps much more surprising is the fact that such set can be Borel. This is true in Cohen's first model.

You can actually replace the infinite Dedekind-finite set by all sort of sets which are "morally smaller than the continuum" (e.g. a set which is the countable union of countable sets, but not countable in itself).

Here is a surprising result, which is a bit in line of what you ask, although not entirely.

Assuming $\sf ZF+DC$, then we do not know of any model in which there is a non-measurable set and $\sf CH$ holds. In particular there is a model in which every set has the Baire Property, but non-measurable sets exist.

This is not entirely what you ask for, because it might be consistent that there are non-measurable sets while every set has size continuum and that $\aleph_1\neq2^{\aleph_0}$; however Shelah's model where every set of reals has the Baire Property (and this contradicts $\sf AC$), but there are non-measurable sets is one where $\aleph_1<2^{\aleph_0}$. The reason that we involve $\sf DC$ in all that is to have a reasonable way of defining a $\sigma$-additive and atomless measure on the reals to begin with.