Holomorphic function has at most countably zeros

Let $f$ be holomorphic in a domain $U$, then $f$ has at most countably zeros.

How can I prove this statement? I have seen it in Sheng Gong explicitly, but without any indications.

If the set $\{z\in U|f(z)=0\}$ were uncountable, then it would have an accumulation point (i.e. not be discrete), and thus you would have $f=0$ by the identity theorem.

Here is a direct proof.

We know zeros of an analytic function are isolated. You shall get the proof in any standard textbook of complex analysis.

So if $a$ and $b$ are two zeros of $f(z)$ with $a \neq b$ you shall get open subsets $A$ and $B$ of $\mathbb{C}$ s.t. $a \in A$ and $b \in B$ with $A \cap B = \phi$

Consider $a_1 + i a_2$ and $b_1 + i b_2$ in $A$ and $B$ with $a_i$, $b_i$ are in $\mathbb{Q}$. So you shall get an one to one correspondence between set of all zeros and a subset of $\mathbb{Q}$.

Thus set of all zeros of the analytic function are countable.