Let $R$ be a Riemann surface that admits a nonconstant 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.

$\begingroup$ What do you mean by "separated" ? $f(z_1)=1$ and $f(z_2)=1$? $\endgroup$ – Loïc Teyssier Dec 18 '18 at 15:26

1$\begingroup$ @LoïcTeyssier We say $f$ separates $a$ and $b$ if $f(a) \neq f(b)$. $\endgroup$ – Jaikrishnan Dec 18 '18 at 16:53

$\begingroup$ If R has finite type, can't you embed R into a ball in C^n and then project to a line through the two points? $\endgroup$ – Autumn Kent Dec 18 '18 at 19:22

1$\begingroup$ @AutumnKent What is your definition of finite type? If $R$ can be embedded into any bounded domain in $\mathbb{C}^n$ then the claim is certainly true. $\endgroup$ – Jaikrishnan Dec 18 '18 at 19:36

$\begingroup$ If it has finitely generated fundamental group. $\endgroup$ – Autumn Kent Dec 18 '18 at 21:54
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 $\sum (1x_n) = \infty$.
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 $H^\infty(R) \cong H^\infty(\Delta)$.)