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Nov 21, 2022 at 8:16 comment added Moishe Kohan @gaga: No, in Sam's example one cannot say this since $\pi_1(X\setminus K)$ is not normally generated by short loops (here $K$ is your compact subset). Also, I do not see why his $X$ cannot not carry a nonconstant bounded subharmonic function.
Nov 21, 2022 at 6:04 comment added gaga Sam's idea and its generalization (Lemma 1) are very useful. By Lemma 1, any nonconstant holomorphic map from Sam's surface $X$ (with a minor modification?) to any closed hyperbolic Riemann surface can be lifted to the universal covering disc on the complement of some compact subset. This gives a nonconstant bounded subharmonic function on $X$, which seems to give a contradiction.
Nov 21, 2022 at 1:57 history answered Moishe Kohan CC BY-SA 4.0