I'm studying linear representations for algebraic groups for the moment. And I kind of got stuck on some theorem. The existence of a finite linear representation makes use of the fact that $V$ is $G$-stable. I got an idea why this should be but I can't find any examples of the opposite.
I messed up with editing, is it possible to remove this question?
I am not completely sure that I understand your question, but maybe this might help:
Consider $V = \mathbb{R}^2$ with $G= GL_{2}(\mathbb{R})$ acting the obvious way. In this example the subspace $U = \mathbb{R} = \{(x,0)\mid x\in \mathbb{R}\}$ is not $G$-stable. You have, for example, that $$ \pmatrix{0 & 1 \\ 1 & 0}\pmatrix{1 \\ 0} = \pmatrix{0 \\ 1}. $$ Here you have a vector in $U$ that is "sent" outside of $U$.