Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein?

Without the axiom of choice, it is possible that there is a vector space $U\neq 0$ over a field $k$ with no nonzero linear functionals.

Let $V$ be the direct sum of countably many copies of $U$, and $W=V\oplus k$.

Then each of $V$ and $W$ embeds in the other, but they are not isomorphic, since $V$ doesn’t have any nonzero linear functionals, but $W$ does.

I don't think there's any restriction on the field $k$, so this answers Variation 1 as well.

There are models of ZF+DC in which every subset of every Polish space has the property of Baire (I can try to add references later, I think to Solovay and Shelah, but these are pretty well known). This implies that every linear map between Banach spaces is continuous.

So we can then take $\ell^\infty$ and $\ell^1$. It is very easy to construct (continuous) linear injections either way: the identity map from $\ell^1$ into $\ell^\infty$, and to go the other way, map $x_n$ to $2^{-n} x_n$.

But if there were a linear isomorphism between them, it would be a homeomorphism, and this is impossible because $\ell^1$ is separable and $\ell^\infty$ isn't.

(As a tie-in to Jeremy's answer, in this model $\ell^1$ is reflexive, and $\ell^\infty / c_0$ is a Banach space with no nonzero linear functionals.)