Is there a compact Riemann surface other than the sphere with an atlas consisting of open subsets biholomorphic to $\mathbb{C}$? Is there a compact Riemann surface other than the sphere which possesses an open subset biholomorphic to $\mathbb{C}$? 



Riemann surfaces with an atlas all of whose open sets are biholomorphic to $\mathbb{C}$?
Using the uniformization theorem, we can prove that the only compact connected Riemann surface admitting an open set biholomorphic to $\mathbb{C}$ is the Riemann sphere $\mathbb{P}^1(\mathbb{C})$. This answers negatively both questions of the OP.


Indeed, let $X$ be a compact connected Riemann surface with an open set $U$ biholomorphic to $\mathbb{C}$. Choose a base point in $U$, and let $\tilde{X}$ be the universal cover of $X$. By the uniformization theorem, $\tilde{X}$ is isomorphic either to $\mathbb{P}^1(\mathbb{C})$, or to $\mathbb{C}$, or to the open unit disk $D$. In the first case we have necessarily $X \cong \mathbb{P}^1(\mathbb{C})$, so let us assume we are not in this case. Since $\mathbb{C}$ is simply connected, the inclusion map $f : \mathbb{C} \to X$ lifts to a map $\tilde{f} : \mathbb{C} \to \tilde{X}$ which is necessarily injective. If $\tilde{X}=D$ then we get a nonconstant bounded entire function, which is impossible. Finally if $\tilde{X}=\mathbb{C}$ then $\tilde{f}$ is a nonconstant entire function, so by Picard's theorem misses at most one value. But $X \cong \mathbb{C}/\Lambda$ and $\tilde{X} \to X$ is far from being injective, which gives a contradiction. (Alternatively, we can use the fact that any injective entire function $\tilde{f}$ is of the form $\tilde{f}(z)=az+b$ with $a \neq 0$, see [this question at MSE](https://math.stackexchange.com/questions/29758/entire-1-1-function).)


**EDIT**. Here is a more elementary argument avoiding the use of the uniformization theorem. The idea was suggested by an attempt of a student of mine.


Let $X$ be a compact connected Riemann surface with an open set $U$ biholomorphic to $\mathbb{C}$. Let $\omega \in \Omega^1(X)$ be a holomorphic $1$-form. The $2$-form $\sigma = i \cdot \omega \wedge \overline{\omega}$ is everywhere non-negative, and is integrable on $X$ since $X$ is compact. It follows that $\sigma$ is integrable on $U$. Let us write $\omega |\_U = f(z) dz$ with $f(z) = \sum\_{n \geq 0} a\_n z^n$ an entire function. Then
\begin{equation\*}
\int\_U \sigma = \int\_{\mathbb{C}} |f(z)|^2 i dz \wedge d\overline{z} = \int\_0^\infty \int\_0^{2\pi} |f(re^{i \theta})|^2 2r dr d\theta.
\end{equation\*}
Now we have
\begin{equation\*}
\int\_0^{2\pi} |f(re^{i\theta})|^2 d\theta = \int\_0^{2\pi} \sum\_{m,n \geq 0} a\_m \overline{a\_n} r^{m+n} e^{i(m-n)\theta} d\theta = 2\pi \sum\_{n \geq 0} |a\_n|^2 r^{2n}.
\end{equation\*}
Assume that $f$ is nonzero, so that there exists an integer $k \geq 0$ such that $a\_k \neq 0$. Then
\begin{equation\*}
\int\_U \sigma \geq 4\pi |a\_k|^2 \int\_0^\infty r^{2k+1} dr
\end{equation\*}
and the last integral is infinite, contradicting the integrability of $\sigma$ on $U$. It follows that $f=0$ and thus $\Omega^1(X)=\{0\}$, so that $X$ has genus $0$ and $X \cong \mathbb{P}^1(\mathbb{C})$.


