Let $X$ be a compact Riemann surface of genus $g>1$ with $G$ a strict open subset of $\mathbb{C}$ and $U=\bigcup_{j}U_j$ strictly a subset of $G$ for $U_j$ compact (for all $j$ such that $\bigcup_{j}U_j$ is compact). Then $\tilde X$ is the open unit disk $\Delta$ which is biholomorphic to $G$ (by the Riemann mapping theorem). Thus the first fundamental group $\pi_1(X)$ is nontrivial, for $X$ is multiply-connected. So $X$ is hyperbolic and we can write it as the quotient of the upper-half plane by a Fuchsian group, i.e. $X=\mathbb{H}/\Gamma.$
Is it true that $X\cong G$ for $\phi:X\to G\subset\mathbb{C}$ biholomorphic?
I am essentially asking if the Riemann surface $X$ of genus $g>1$ is diffeomorphic to an open subset of $\mathbb{C},$ $G\cong\Delta.$
Thanks in advance!
In a somewhat trivial, but perhaps significant, way, the answer is "no": compact connected Riemann surfaces of genus >1 are not diffeomorphic to open subsets of $\mathbb C$". Namely, open subsets of $\mathbb C$ are not compact. If we try to dodge this by taking closures, then the uniformization theorem will require that we identify some points on the boundaries, so the map (from a non-Euclidean polygon to the Riemann surface) will definitely not be one-to-one.
Possibly the genuine intent of the question is somewhat different from this, or can be refined to take into consideration such "trivial counter-examples", to approach the real issue?