Let $F(x,y)=(e^x\cos y, e^x\sin y)$, show that $F$ isn't invertible on all of $\mathbb R^2$, although it's locally invertible everywhere.
It's obvious that $F$ is locally invertible everywhere, because $$J_F(x,y)=\pmatrix{e^x\cos y & -e^x\sin y\\ e^x\sin y & e^x\cos y}$$ so the determinant of the jacobian matrix is $$\Delta_F(x,y)=e^{2x}\cos^2x+e^{2x}\sin^2x=e^{2x}\ne0$$ Hence $F$ is locally invertible everywhere, but i can't prove that $F$ isn't invertible.
You have already done the hard part of showing locally invertible. To show that it is not invertible, it suffices to show that it is not one-to-one. (Why?)
For that, simply note that $F(0, 0) = F(0, 2\pi)$ but $(0, 0) \neq (0, 2\pi)$.