Let $W$ be a vector space. Let $U$ be a subspace and $V$ be a subgroup of $W$, and assume that $W$ is the direct sum (as groups!) of $U$ and $V$. Does it follow that $V$ is a subspace, such that in fact $W=U\oplus V$ as vector spaces? (In case a counterexample is found, i'd like to see if the answer is the same for finite and infinite dimensional vector spaces?)
Put another way, if $U$ and $V$ are complementary subgroups and $U$ is closed under scalar multiplication, does it follow that $V$ is also closed under scalar multiplication?
I can't see a particular reason why this should be the case, but at the same time I can't seem to find a counterexample, so here we are.... As a related example, considering $\Bbb C$ as a complex vector space, we have $\Bbb C = \Bbb R \oplus i\Bbb R$ as groups, but neither of these subgroups are subspaces, so it's not applicable.
To do things the proper way, let me flesh out the solution in the comments as an answer.
The idea suggested is to consider a field extension $K/F$ and a vector space $W$ over $K$ (which thereby also can be viewed as a vector space over $F$). Then choose $U$ as a $K$-linear subspace and $V$ as an $F$-linear complement. In general (probably always?), there are many choices for $V$ that are not $K$-linear, which means we get a solution to the problem.
Concrete example: Take $W=\Bbb C^2$, $U=\{(z,0)\mid z\in\Bbb C\}$, $V=\{(\alpha, \alpha+i\beta)\mid \alpha,\beta\in\Bbb R\}$.