I was looking at this question: Can a $0$-dimensional analytic subset of a compact complex variety has infinite many points?. The answer says that the answer to my title question is yes, but appears to only prove it when $M=\mathbb{P}^1$. I was wondering how to prove it in the general case, if it is actually true. Here are the definitions I am using.
Definition of subvariety and dimension: Let $M$ be a complex manifold and $V\subset M$ a closed subset. Then $V$ is called an analytic subvariety of $M$ if for all $p\in V$, there is an open neighbourhood $U\subset M$ of $p$ such that $$U\cap V=Z(f_1,\dots,f_n)=\{x\in U\ :\ f_1(x)=\dots=f_n(x)=0\}$$ where each $f_i\in\mathcal{O}(U)$. The dimension of $V$ is the largest integer $k$ such that there is a neighbourhood $U\subset M$ such that $U\cap V$ is a complex manifold of dimension $k$.
Actual statement I would like to prove/disprove: Let $V\subset M$ be a $0$-dimensional analytic subvariety of a compact complex manifold $M$. Then $V$ is finite.