How can I argue that Lie derivative is not a connection? I am reading Lee's book of riemannian geometry and he asks to show that Lie derivative of two vector fields on a riemannian manifold is not a connection.
How can I argue that this is true?
He also asks to show that there is a vector field $V$ on $\mathbb{R}^2$ such that $V$ vanishes on the $x$-axis but $\mathcal{L}_{\partial_x}V$ does not.
This was a confusion to me too.
I can take for example: $V = x\partial_x.$
Then $$[\partial_x,x\partial_x]f = \partial_x(x\partial_xf) - x\partial_x(\partial_xf) = \partial_xf + x\partial^2_{xx}f - x\partial_{xx}^2f = \partial_xf.$$
Then $$[\partial_x,x\partial_x] = \partial_x.$$
And that is a possible solution.
Is this right?

Your example for the second problem would be OK, except that I suppose your $V$  vanishes on the $y$-axis instead of the $x$-axis.
Your example is also the sort of thing you should think about to solve your first problem. The axiom $\nabla_{fX} Y = f \nabla_X Y$ for a connection is actually equivalent to saying that, for fixed $p \in M$, the map $Y \mapsto (\nabla_XY)(p)$ only depends (linearly) on $X(p)$. This equivalence follows from the divisibility properties of smooth functions (if $f : \mathbb{R}^n \to \mathbb{R}$ is smooth and vanishes at the origin, then one can write $f$ as a finite sum $\sum f_i g_i$ where all functions are smooth and the $f_i$ vanish at the origin). 
The conclusion is that connections have the following property: $(\nabla_XY)(p)$ must be zero if $X(p)=0$. The Lie derivative does not have this property, as you can discover by thinking about your example. 
