= Solution
The <uniformization theorem> says that every <simply connected> <Riemann surface> is <biholomorphic> to the <Riemann sphere>, the <complex plane>, or the <unit disc>. Its consequence for a <compact Riemann surface> of <genus> at least two is that the <universal cover> is the <unit disc>, equivalently the <hyperbolic plane>.
To explain the restriction, a spherical <universal cover> would make the <fundamental group> finite. Since a finite <covering map> multiplies <Euler characteristic>, $2=m(2-2g)$ is impossible for $g\geq2$. If the <universal cover> were the <complex plane>, its <deck transformations> would be <biholomorphisms> $z\mapsto az+b$. A nonidentity such map with $a\ne1$ has a fixed point, impossible for a <deck transformation>. Thus the <deck transformation group> would consist of translations. A <discrete subgroup> of translations with <compact> <quotient space> is a rank-two <Euclidean lattice>, whose <quotient space> is a <torus>, again incompatible with $g\geq2$.
On the <unit disc>, the <Poincare metric on a Riemann surface>
$$
ds^2=\frac{4|dz|^2}{(1-|z|^2)^2}
$$
has <Gaussian curvature> $-1$ and is invariant under every <biholomorphism> of the <unit disc>. It therefore descends through the <deck transformation group> to a <hyperbolic metric> on $X$. Its local expression is a positive multiple of $|dz|^2$ in every holomorphic <coordinate chart>, so it is compatible with the given <complex structure>. In fact, this <hyperbolic metric> is unique within that <conformal class>.
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