Stronger Liouville theorem "Every bounded function that is holomorphic on $A$ is is constant."
For which $A\subseteq\mathbb{C}$ is this true?
Are there well-known examples of unbounded sets $A\subseteq\mathbb{C}$ on which there are non-constant bounded holomorphic functions?
Later edit: My striking through the second question was meant only to de-emphasize it.  Feel free to post further on it if you wish.  Some of the examples posted in response to it were already well known to me; I'd have thought of them if my attention had been on the second question rather than the first.
I'm envisioning a couple of possibilities: (1) Various other sorts of sets $A$ will be mentioned in answers; and (2) An answer will say that some nice theorem says this is true of a set $A$ if and only if whatever, where "whatever" is something non-trivially different from a tautologous "if and only if every bounded holomorphic function on $A$ is constant", and maybe "whatever" is somehow elegant or at least simple.
 A: For any $A$ whose complement contains a neighborhood of a point $z_o$, the holomorphic function $1/(z-z_o)$ is bounded on $A$.
A: On the affirmative side you have that if $B$ is a discrete subset of $\mathbb{C}$ (meaning it has no accumulation points) then every bounded holomorphic function $f: \mathbb{C}\setminus B \to \mathbb{C}$ must be constant. This is because every "singularity" at $b\in B$ is removable (here you use that $B$ is discrete) and so $f$ extends to a bounded entire function.  
A: Look at the exponential function on the left half-plane.  If $a\le 0$,
$$|\exp(a + ib)| = e^a \le 1.$$
A: If $A$ is connected and open, by the uniformization theorem it is covered by either $\mathbb{C}$, the open disk, or the Riemann sphere. The third possibility does not occur in this case so we are reduced to the first two. In the first case any bounded entire function on $A$ pulls back to a bounded entire function on $\mathbb{C}$ so we reduce to the usual Liouville's theorem. In the second case the open disk clearly admits a non-constant bounded entire function to $\mathbb{C}$, namely the obvious inclusion. In particular, if $A$ is simply-connected and covered by the open disk then it is in fact biholomorphic to the open disk (so for example any open half-space has this property). 
A: There are examples to your second question. Let $A$ to be the upper-half-plane, and consider the canonical conformal map from the upper-half-plane to the unit disc given by
$$
F(z) = \frac{i-z}{i+z}
$$
