First of all, note that even with the axiom of choice there isn't necessarily a unique complement.
For example, take $V=\Bbb R^2$ and $\newcommand{\sp}{\operatorname{span}}W=\sp\{(0,1)\}$. Then both $\sp\{(1,1)\}$ and $\sp\{(1,0)\}$ are direct complements of $W$,since their internal direct sum with $\sp\{(0,1)\}$ give $V=\Bbb R^2$, however neither is equal to the other.
In the case where we want an orthogonal complement, then we no longer talk about vector spaces, but rather we talk about inner product spaces.
Finally, the axiom of choice is equivalent to the statement that direct complements exist. That is, if $V$ is a vector space and $W$ is a subspace then there is some $U$ such that $W\cap U=\{0\}$ and $\sp(W\cup U)=V$. This means that when the axiom of choice fails then there is some vector space $V$ and a subspace $W$ such that $W$ has no direct complement, every subspace will either intersect nontrivially with $W$ or their sum won't be the entire space.
However the failure of the axiom of choice is as non-constructive as the axiom of choice itself. So we can't really construct an example. We can, though, use the fact that there is a family of non-empty sets which doesn't admit a choice function, and by analyzing the proof of the above equivalence construct a vector space and a subspace without a direct complement.
If one only wishes to produce a consistency result (i.e. construct a particular example), then one can use Läuchli's construction (and by extension the work I did in my masters) and show that it is possible to have a vector space which is not finitely dimensional, but every proper subspace has a finite dimension. Clearly, in such vector space there is no direct complement to any proper subspace.
Finally, now to answer the question in the title. Take a model where there is such a Läuchli space, $W$ and consider $V=W\oplus W$. Then the subspace $W\oplus\{0\}$ has a unique direct complement.
