Why does a holomorphic differential has $2g-2$ zeros? If $X$ is a compact Riemann surface, then any holomorphic differential on $X$ has $2g-2$ zeros.
I would like to know how to prove this. If possible, without some "heavy machinery" like divisors and the Riemann-Roch theorem, which I don't understand very well yet. I really think that we can use the Riemann-Hurwitz to do this, but I am not sure how.

If you know what is the Euler characteristic, this is easy. A holomorphic form is a section of the cotangent bundle, whose Euler characteristic is $-(2-2g)$ (this bundle is the dual of the tangent bundle), and all zeroes of this section have to be counted with a $+1$, as we choose an holomorphic form. 
