= Solution
The <uniformization theorem> states that every simply connected Riemann surface is conformally equivalent to exactly one of the <Riemann sphere> $\mathbb C_\infty$, the <complex plane> $\mathbb C$, or the open unit disc $\mathbb D$. Equivalently, every connected Riemann surface is a quotient of one of these by a free properly discontinuous group of conformal automorphisms.
For $\mathbb C_\infty$, every nonidentity Möbius transformation has a fixed point. A free action is therefore trivial, so the only surface uniformized by the sphere is
$$
\boxed{\mathbb C_\infty.}
$$
Every conformal automorphism of $\mathbb C$ has the form $z\mapsto az+b$. If $a\ne1$, it has a fixed point, so a group acting freely consists only of translations. A discrete subgroup of $(\mathbb C,+)$ has rank zero, one, or two. The corresponding quotients are
$$
\boxed{\mathbb C,\qquad
\mathbb C/\omega\mathbb Z\cong\mathbb C^*,\qquad
\mathbb C/\Lambda,}
$$
where $\Lambda$ is a rank-two lattice and $\mathbb C/\Lambda$ is a complex torus. This proves the completeness of the <Riemann surfaces uniformized by the complex plane>.
Now let $D\subseteq\mathbb C$ be a domain whose complement contains two distinct points. Its universal cover cannot be the sphere because no nonconstant holomorphic map from a compact Riemann surface has image in the noncompact proper domain $D$. If its universal cover were $\mathbb C$, the covering map followed by $D\hookrightarrow\mathbb C$ would be a nonconstant entire function omitting the two chosen complementary points, contradicting the <Little Picard theorem>. The uniformization theorem therefore leaves only
$$
\boxed{\widetilde D\cong\mathbb D,}
$$
which is the <plane domain with two omitted points is hyperbolic> result.
Finally let $j:\mathbb C^*\hookrightarrow R$ be a holomorphic embedding with $R$ compact. At each puncture zero and infinity, an essential singularity would, by the <Great Picard theorem> in a coordinate chart, force values to be taken repeatedly and contradict injectivity. Thus $j$ extends holomorphically to
$$
J:\mathbb C_\infty\to R.
$$
A generic point of the open image $j(\mathbb C^*)$ has exactly one preimage, so $J$ has degree one. A degree-one holomorphic map between compact Riemann surfaces is a biholomorphism; equivalently, the <Riemann-Hurwitz formula> rules out branching and gives the same conclusion. Therefore the <compact Riemann surface containing an embedded punctured plane> is
$$
\boxed{R\cong\mathbb C_\infty.}
$$
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