As has been pointed out, the inequivalence of the three is elementary.

The original proofs of Koebe and Poincare were by means of harmonic functions, i.e. the
Laplace equation ${\Delta}u = 0$. This approach was later considerably streamlined by means of Perron's method for constructing harmonic functions. Perron's method is very nice, as it is elementary (in complex analysis terms) and requires next to no topological assumptions.
A modern proof of the full uniformization theorem along these lines may be found in the book "Conformal Invariants" by Ahlfors.

The second proof of Koebe uses holomorphic functions, i.e. the Cauchy-Riemann equations, and
some topology.

There is a proof by Borel that uses the nonlinear PDE that expresses that the Gaussian curvature is constant. This ties in with the differential-geometric version of the Uniformization Theorem: Any surface (smooth, connected 2-manifold without boundary) carries a Riemannian metric with constant Gaussian curvature. (valid also for noncompact surfaces).

There is a proof by Bers using the Beltrami equation (another PDE).

For special cases the proof is easier. The case of a compact simply connected Riemann surface can be done by constructing a nonconstant meromorphic function by means of harmonic functions, and this is less involved than the full case. There is a short paper by Fisher, Hubbard and Wittner where the case of domains on the Riemann sphere is done by means of an idea of Koebe. (Subtle point here: Fisher et al consider non-simply connected domains on the Riemann sphere. The universal covering is a simply connected Riemann surface, but it is not obvious that it is biholomorphic to a domain on the Riemann sphere, so the Riemann Mapping Theorem does not apply).

The Uniformization Theorem lies a good deal deeper than the Riemann Mapping Theorem.
The latter is the special case of the former where the Riemann surface is a simply connected domain on the Riemann sphere.

I decided to add a comment to clear up a misunderstanding. The theorem that a simply connected surface (say smooth, connected 2-manifold without boundary) is diffeomorphic to the plane (a.k.a. the disk, diffeomorphically) or the sphere, is a theorem in topology, and is *not* the Uniformization Theorem. The latter says that any simply connected Riemann surface is *biholomorphic* (or conformally equivalent; same in complex dimension $1$) to the disk, the complex plane or the Riemann sphere.

But the topology theorem is a corollary to the Uniformization Theorem. To see this, suppose $X$ is a simply connected (smooth etc.) surface. Step (1): Immerse it in $\mathbb{R}^3$ so as to miss the origin. Step (2): Put the Riemann sphere (with its complex structure!) in $\mathbb{R}^3$ in the form of the unit sphere. Step (3): For every tangent space $T_pX$ on $X$, carry the complex structure $J$ from the corresponding tangent space on the Riemann sphere by parallell transport (Gauss map) to $T_pX$. This is well-defined by choosing a basepoint and recalling that $X$ is simply connected. Step (4): Presto! $X$ is now a Riemann surface (it carries a complex structure), so it is biholomorphic to the disk or the plane or the Riemann sphere, thus diffeomorphic to one of the three.

Of course, I have glided over the question of immersing the surface in 3-space, because this is topology. Actually, I vaguely recall that there is a classification of noncompact topological surfaces by Johannsen (sp?), and no doubt the topological theorem would immediately fall out of that.