Prove that closed orientable Riemann surfaces are hyperbolic pecicely when their genus is greater than $1$ I roughly know that I will have to turn to the Uniformization Theorem for this and probably also to the Gauss-Bonnet Theorem. But I'm having trouble making it precise.
Here's what I can tell you:
$\scriptsize\text{(From Wikipedia)}$

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*(Uniformization Theorem) Every simply-connected Riemann surface is conformally equivalent to $\mathbb {\hat C}$, $\mathbb C$ or $\mathbb D(\cong\mathbb H$).

*(Gauss-Bonnet) Take $\Sigma_g$ a closed orientable Riemann surface and $k$ its curvature, then $\iint_{\Sigma_g}k\;  dA<0\iff g>1$.

I want to be able to deduce that, for $g>1$, every $\Sigma_g\cong \mathbb D/\Gamma$, where $\Gamma$ is some discrete group, and that it thus makes sense to study fundamental polyons of $\mathbb D$.
I feel like I need something more than just Uniformization and Gauss-Bonnet for this. Especially since the Uniformization Theorem only speaks of simply-connected Riemann surfaces.
I need to know what extra steps to take when I want to relate $\Sigma_g$'s (not simply-connected) to $\mathbb D/\Gamma$'s ($\neq\mathbb D$).

Edit
The comments gave me an idea:
Every $\Sigma_g$ has a universal cover $p:U\to\Sigma_g$. I can say $\Sigma_g=U/p$ and give it the induced metric $d_{\Sigma_g}(x,y)=\inf d_U(p^{-1}x,p^{-1}y).$ Then I use Gauss-Bonnet to say $$\iint_{\Sigma_g}k \;dA=\iint_S k \; dA<0,$$ where $S$ is a single sheet above $\Sigma_g$. $U$ is the sum of these sheets, so $\iint_U k\; dA<0$. By the Uniformisation Theorem  $U$ must be conformally equivalent to $\mathbb D$.
Is this correct?
 A: You start with a compact connected Riemann surface $\Sigma$ of genus $\ge 2$; lift its complex structure to the universal cover $X=\tilde\Sigma$; then the group $G$ of covering transformations will act conformally on $X$. By the Uniformization Theorem, $X$ is conformal either to the open unit disk $D$ or to the complex plane ${\mathbb C}$. In either case, $G$ embeds as a discrete subgroup (acting freely, since this is a covering action) of the unit disk or of the complex plane. Let's rule out the latter case. From a Complex Analysis class, you already know that the group of conformal automorphisms of ${\mathbb C}$ consists of complex-affine transformations $z\mapsto az+b$. But, unless $a=1$, every such transformation has a fixed point in  ${\mathbb C}$. Hence, $G$ acts as a group of translations and, hence, is abelian. Now, you use your knowledge of Algebraic Topology (or Differential Topology) and check that in this case ${\mathbb C}/G$ is the torus, i.e. has genus 1, which is a contradiction. (For instance, note that ${\mathbb C}$ has a unit vector field invariant under the group of translations, which implies that ${\mathbb C}/G$ has zero Euler characteristic.) There are other ways to argue at this point as well (you can use Gauss-Bonnet as well since ${\mathbb C}/G$ has flat metric).
Now, you know that $X$ is conformal to the unit disk $D$, hence, $G$ acts on $D$ as a group of conformal automorphisms. It is a corollary of Schwarz Lemma that $Conf(D)$ equals the group of linear-fractional automorphisms of $D$, and, hence, preserves the hyperbolic metric. Therefore, $\Sigma$ is conformal to the hyperbolic surface $D/G$.  
One more thing: Riemann surfaces are automatically oriented (the same is true for complex manifolds manifolds in all dimensions), but one needs to assume that the surface is connected.
