Let $X$ be a compact connected Riemann surface. Let $\{h_1(z),\ldots,h_g(z)\}$ be a base of the space of holomorphic forms on $X$ in a local map centered in $p\in X$. Let’s assume that we know that $p$ is a Weierstrass point iff the Wronskian
$$W(z):=\begin{vmatrix}h_1(z) & h_2(z) & \ldots & h_g(z) \\h_1’(z) & h_2’(z) & \ldots & h_g’(z) \\ \vdots & \vdots & & \vdots \\h_1^{(g-1)}(z) & h_2^{(g-1)}(z) & \ldots & h_g^{(g-1)}(z)\end{vmatrix}$$
verifies $W(0)=0$ (we can take this as the definition of a Weierstrass point for the purposes of this question) (I asked for a proof of this equivalence here).
Let’s also assume that we know that $W(z)$ isn’t identically null on any local chart.
I’d like to check if the following proof (that I came up with) of the fact that the number of Weierstrass points in $X$ is finite is correct, or if it could be made more rigorous:
Since all the $h_i(z)$ are holomorphic functions, then $W(z)$ is holomorphic too. But then the set of zeros of $W(z)$ must be discrete, or else it would contain an accumulation point, and by the identity principle $W(z)$ would be identically null. We also know it is closed since $W(z)$ is continuous, so it is closed and discrete. But every closed and discrete subset of a compact is finite, so the set of zeros of $W(z)$ is finite… and since it must contain the set of Weierstrass points of $X$, this set is finite too.
I think you are using the right concepts from complex analysis and topology. However, you should be careful that $W(z)$ is only defined locally, and cannot be a holomorphic function on all of $X$, because all holomorphic constant function on a compact Riemann surface are constant.
So what you write asserts that locally the set of Zariski points is discrete and closed, hence it is discrete and closed globally. On a compact space, discrete and closed implies finite.