It is a known result by A. Blass that the axiom of choice is equivalent to the assertion that every vector space has a basis. (Rubin's Equivalents of the Axiom of Choice: form B)

It is also known that the axiom of choice equivalent to the existence of a field $F$ such that for every vector space $V$ over $F$ has the property that if $S\subseteq V$ is a subspace of $V$ then there is $S'\subseteq V$ such that $S\oplus S'=V$.

In this case, assuming the axiom of choice does not hold, we have a vector space without a basis somewhere in the model, and over every field there is a vector space $V$ that has a subspace without a complement.

**My question**: Can we derive from the absence of choice that there will be a vector space (over some field? any field?) such that for any nontrivial subspace $S$ of $V$ there is no subspace $S'$ of $V$ such that $S\oplus S'=V$?

My initial instinct was to try and take a vector space without a basis and start decomposing it, somehow deriving if we stop at a successor stage we have found such subspace, and if we happened to hit a limit stage with an empty subspace then we can construct a basis. I think I overlooked something because I couldn't finish my argument.

*Edit:* Just to be clear, I am not looking whether or not this is consistent with ZF, but rather if the negation of choice implies there exists such vector space.