I am trying to prove that if a meromorphic function is constant on an open subset $U$ then it is constant in all its domain $\Omega$. My idea is:
Let $F$ be meromorphic in $\Omega$. Then we have $F=\frac{f}{g}$ with $f$ and $g$ analytic. Since $F$ is holomorphic in $U$ we have $f-gK=0$. Now since $f-gK=0$ is holomorphic in $U$ by the identity theorem we have $f-gK=0$ in $\Omega$ and so $F=\frac{f}{g}=K.$
Is my proof correct, or I am missing something?
My main concern here is to know if every meromorphic function is of the form $F=\frac{f}{g}$ with $f$ and $g$ analytic.
Your proof is correct, but I think you can do it simpler.
A meromorphic $F$ function on a domain $\Omega$ is a function that is holomorphic on all of $\Omega$ except for a set $A$ of isolated singularities which are poles of the function. The set $B = \Omega \setminus A$ is open and connected and contains $U$, $F \mid_B$ is holomorphic, hence the identity theorem shows that $F \mid_B$ must be constant. This shows additionally that all possibly existing singularities are removable, thus $A$ must be empty.