Usually the domain of a complex function is $\mathbb C\backslash\{z\in\mathbb C~:~z \text{ is a singularity of } f \}$ So I guess it must always be $\mathbb C\backslash\{\text{a set of points}\}$. Moreover it's usually a finite or at most countable set of points.
So the domain doesn't need to be open because it can be the infinite intersection $\bigcap\limits_{i\in A}{}\mathbb C\backslash\{z_i\}$ and $A$ counld be infinite. For instance $\bigcap\limits_{n\in\Bbb N}{}\mathbb C\backslash\{{1\over n}\}$ is not open.
Can the domain be closed?
Is there a complex function that is not defined on a connected set that is more than a segment?
What if $f$ is holomorphic/meromorphic?
First of of all, asserting that the domain of a function $f$ is the complement of the set of singularities of $f$ makes no sense, because it is clearly a circular defintion.
It is just a matter of definition. Often, but not always, , the domain of a holomorphic function is open by definition.
On the other hand, if $r\in(0,1]$ and if I define$$\begin{array}{rccc}f\colon&D(0,r)&\longrightarrow&\mathbb C\\&z&\mapsto&\displaystyle\sum_{n=0}^\infty z^n,\end{array}$$then the domain of $f$ is $D(0,r)$. This has nothing to do with singularities. And, yes, the domain can also be $\overline{D(0,r)}$ (if you are not assuming that the domain is open by definition), unless $r=1$.