The sphere, plane and disc are pairwise biholomorphically distinct

The plane $\mathbb{C}$ and the disc $\mathbb{D}$ are plane domains, and each carries the one-chart atlas given by the identity map with trivial transition; hence $\mathbb{C}$ and $\mathbb{D}$ are Riemann surfaces. Both are convex subsets of ${\mathbb{R}}^2=\mathbb{C}$: for $\mathbb{D}$ and $z,w\in \mathbb{D}$, $t\in [0,1]$, one has $\mathrm{\mid }(1-t)z+tw\mathrm{\mid }\le (1-t)\mathrm{\mid }z\mathrm{\mid }+t\mathrm{\mid }w\mathrm{\mid }<1$, and the case of $\mathbb{C}$ is immediate.

The sphere $\widehat{\mathbb{C}}$ is a Riemann surface: its standard charts have holomorphic transition maps [F2], and $\mathrm{\Sigma }$ is a homeomorphism onto $S^2$ [F3], so $\widehat{\mathbb{C}}$ is Hausdorff and second countable (as a metrizable space) and nonempty; moreover $S^2$ is path-connected because it is simply connected [F4, F5], and a continuous image of a path-connected space is path-connected, so $\widehat{\mathbb{C}}$ is connected.

Neither $\mathbb{C}$ nor $\mathbb{D}$ is compact. The discs $D(0,n)$, $n\ge 1$, cover $\mathbb{C}$, but any finitely many of them lie in $D(0,N)$ with $N$ the largest index and omit points of large modulus, so they do not cover $\mathbb{C}$. Likewise the discs $D(0,1-1\mathrm{/}n)$, $n\ge 2$, cover $\mathbb{D}$, but any finitely many lie in $D(0,1-1\mathrm{/}N)$ with $N$ the largest index and omit points of modulus between $1-1\mathrm{/}N$ and $1$. By the definition of compactness, neither space is compact.

The plane and the disc are simply connected, being nonempty convex subsets of ${\mathbb{R}}^2$ with their Euclidean topology.

The sphere $\widehat{\mathbb{C}}$ is simply connected. Since $S^2$ is simply connected for $n=2$ [F4], every loop in $S^2$ based at a given point represents the identity class [F5]. Let $x\in \widehat{\mathbb{C}}$ and let $\alpha$ be a loop at $x$; then $\mathrm{\Sigma }\circ \alpha$ is a loop at $\mathrm{\Sigma }(x)$ in $S^2$, so $[\mathrm{\Sigma }\circ \alpha ]$ is the identity of ${\pi }_1(S^2,\mathrm{\Sigma }(x))$, while by functoriality [F6] the homomorphism ${\mathrm{\Sigma }}_{\ast }:{\pi }_1(\widehat{\mathbb{C}},x)\to {\pi }_1(S^2,\mathrm{\Sigma }(x))$ has inverse $({\mathrm{\Sigma }}^{-1}{)}_{\ast }$, because $({\mathrm{\Sigma }}^{-1}\circ \mathrm{\Sigma }{)}_{\ast }={\mathrm{id}}_{\ast }=\mathrm{id}$. Hence $[\alpha ]=({\mathrm{\Sigma }}^{-1}{)}_{\ast }([\mathrm{\Sigma }\circ \alpha ])$ is the identity, so ${\pi }_1(\widehat{\mathbb{C}},x)$ is trivial for every $x$; combined with the nonemptiness and path-connectedness from step 1.2, the sphere is simply connected.

The sphere $\widehat{\mathbb{C}}$ is compact: ${\mathrm{\Sigma }}^{-1}:S^2\to \widehat{\mathbb{C}}$ is continuous [F3], and $S^2$ is compact, being a Euclidean sphere [F10]; a continuous image of a compact set is compact [F11].

The plane is not biholomorphic to the disc: if $f:\mathbb{C}\to \mathbb{D}$ were a biholomorphism of complex domains, then $f$ is holomorphic and bijective [F13], and regarded as a map into $\mathbb{C}$ it is entire with $\mathrm{\mid }f(z)\mathrm{\mid }<1$ for all $z$; by Liouville's theorem [F8] $f$ would be constant, and a constant map is not bijective, a contradiction.

The sphere is not biholomorphic to the plane: if $f:\widehat{\mathbb{C}}\to \mathbb{C}$ were a biholomorphism, it would be a continuous bijection [F12], and then $\mathbb{C}=f(\widehat{\mathbb{C}})$ would be a continuous image of the compact space $\widehat{\mathbb{C}}$, hence compact [F11], contradicting step 1.3.

The sphere is not biholomorphic to the disc: the same argument with $\mathbb{D}$ in place of $\mathbb{C}$ shows that a biholomorphism $\widehat{\mathbb{C}}\to \mathbb{D}$ would make the noncompact space $\mathbb{D}$ a continuous image of the compact sphere.

Steps 1.2, 2.1 and 2.2 show that $\widehat{\mathbb{C}}$, $\mathbb{C}$ and $\mathbb{D}$ are simply connected Riemann surfaces, and steps 3.1, 3.2 and 2.4 show that no two of them are biholomorphic. No choice principle is used: the only compactness arguments use the explicit countable covers displayed in step 1.3, and the rigidity argument is Liouville's theorem.