Let $R$ be a Riemann surface that admits a non-constant bounded holomorphic function. Then is it true that any two points of $R$ can be separated by a bounded holomorphic function? This is easy to see when $R$ is a planar domain. 



Bounded holomorphic functions on a Riemann surface separating points
The answer is no.


See


Stanton, Charles M., Bounded analytic functions on a class of open Riemann surfaces. Pacific J. Math. 59 (1975), no. 2, 557–565.


Stanton shows that a branched cover $R$ of the disk is separated by $H^\infty(R)$ if and only if the branch points lie over a set that is the zero set of a Blaschke product. So you can build an example like this:


Take a $2$-fold branched cover $\pi:\Delta\_2 \to \Delta$ of the unit disk $\Delta$ with branched points above a sequence $x\_1, x\_2, \ldots$ with $0 <x\_1 < x\_2 < \cdots < 1$ tending to $1$ and with $\sum (1-x\_n) = \infty$. 


If $f$ is a bounded analytic function on $\Delta\_2$ define $h(z) = (f(z\_1) - f(z\_2))^2$, where $\pi^{-1}(z) = \{z\_1,z\_2\}$. The function $h$ is an analytic function on the disk, and the Blaschke theorem and the choice of $x\_n$ forces $h$ to be identically zero.


It seems like the characterization of Riemann surfaces $R$ separated by $H^\infty(R)$ is still an open problem.


(My first answer cited example 2 on page 241 of 


Nakai, Mitsuru, Valuations on meromorphic functions of bounded type. 
Trans. Amer. Math. Soc. 309 (1988), no. 1, 231–252.


This is a variation of the above that is not a branched disk and yet $H^\infty(R) \cong H^\infty(\Delta)$.)


