Question:
Let $M$ denote a two-dimensional, smooth, compact Riemannian manifold (without boundary). If $\phi:\mathbb{R}\times M \rightarrow M$ is a smooth flow, then prove that $\phi$ is not an Anosov flow.
Motivation:
I am new to the field of dynamical systems. It seems to be implicit in the literature that Anosov flows do not exist on two dimensional manifolds (see http://www.sciencedirect.com/science/article/pii/S0022039605002767) but I would like to know the proof.
My current understanding:
Assume $\phi$ is Anosov and argue for a contradiction. There is a splitting of the tangent space $T_x M = E_x^s \oplus E_x^u \oplus E_x^c$ at each $x\in M$, where $E_x^c$ is the one-dimensional subspace spanned by the direction of the flow, and the stable and unstable spaces have dimension independent of $x$. Thus, without loss of generality, suppose that $E_x^u = \{0\}$ and $\text{dim}(E_x^s)=1$. The differential of the vectors $v$ in $E_x^s$ satisfy $\|d\phi_tv\| \lesssim e^{-\gamma t} \|v\|$ for all $t\geq 0$ and some $\gamma >0$. I guess that I now need to use the fact that $M$ is compact to obtain a contradiction but I don't see how to proceed. Any help would be much appreciated.

If M is the 2-torus then every hyperbolic automorphism gives rise to an Anosov diffeomorphism. In fact the most known and used example is Arnold's CAT map, given by the hyperbolic matrix $A=\begin{pmatrix}2&1\\1&1\end{pmatrix}$.
In this case you have $\mathbb{R}^2$ being the direct sum of a one-dimensional expanding subspace and a one-dimensional contracting subspace. The center-space is zero dimensional in this case.