Let’s assume that every holomorphic $1$-form on a compact and connected Riemann surface $X$ of genus $g=2$ has two zeroes (with multiciplity). I want to prove that there exist two holomorphic $1$-forms $\omega_1$, $\omega_2$, on $X$ such that their sets of zeroes are disjoint.
This is what I thought: I know that the dimension of the space of holomorphic $1$-forms on $X$ is $g=2$, and I also know that if two holomorphic $1$-forms share the same zeroes, then they must be proportional. Then, there must be some $\omega_1$, $\omega_2$ that don’t share at least one of their zeroes, or else the dimension of the space of holomorphic $1$-forms on $X$ would be $1$ (they’d all be proportional to each other).
What if they have just a zero in common? Then I think they would define a meromorphic function $\omega_1/\omega_2$ on $X$ with exactly one pole and one zero. But then $X$ would be isomorphic to the Riemann sphere as Riemann surfaces, which is impossible since they have different genus.
Is this idea correct?