Statement
Let $f:\mathbb \Omega \to \mathbb C$ be a holomorphic function, $f \neq 0$ ($\Omega$ is a region, i.e., an open, nonempty, connected set). Prove that in every compact subset $K$ of $\Omega$, the set of zeros of $f$ is finite.
I've read in Stein's textbook a proof of the statement "Suppose $f$ is a holomorphic function in a region $\Omega$ that vanishes on a sequence of distinct points with a limit point in $\Omega$. Then $f$ is identically zero."
From that statement, one can easily deduce that the zeros of $f$ are isolated. I've written a proof of the original statement using that fact. I would like to know if my proof is correct and also to encourage to give an answer to anyone who has an alternative solution (or my corrected solution, should some step of mine be wrong).
Proof of statement
The proof is by contradiction. I'll denote $S=\{z \in K : f(z)=0\}$. Define $\{F_j\}_{\{j \in \mathcal J\}}$ to be $\{F_j\}_{\{j \in \mathcal J\}}=\bigcup_{\{i \in \mathcal I\}} B(z_i,\delta_i) \cup \ C$, where $C$ is an open cover for $S^c$, and $B(z_i,\delta_{i})$ is an open ball centered at $z_i$ with a radius chosen such that $z \not \in B(z_i,\delta_{i})$ for $z \in S$ different from $z_i$, for each $z_i \in S$ . It is clear that $\{F_j\}_{\{j \in \mathcal J\}}$ is an open cover for $K$. Now, since $K$ is compact, there exists a finite subcover $\{F_{j'}\}_{\{j' \in \mathcal J' \subset J\}}$, with $J'$ finite. By the way the original cover was defined, the subcover must be of the form $\{F_{j'}\}_{\{j' \in \mathcal J' \subset J\}}=\bigcup_{\{i' \in \mathcal I'\}} B(z_{i'},\delta_{i'}) \cup \ C'$, where $I'$ is a finite subset of $I$ and $C'$ is a finite subcover extracted from $C$.
Now choose $z \in S$ with $z \neq z_{i'}$ for all $i' \in I'$ (we can do this since the set of zeros is supposed to be infinite). By the way the open balls were chosen, $z \not \in \bigcup_{\{i' \in \mathcal I'\}} B(z_{i'},\delta_{i'})$, but $z \in K=\bigcup_{\{i' \in \mathcal I'\}} B(z_{i'},\delta_{i'}) \cup \ C'$, so $z$ must be in $C'$, but this is clearly absurd since $C'$ is a cover of $S^c$. The absurd comes from the assumption that $S$ is infinite, then $S$ must be finite.
Best Answer
Let $K \subset \Omega$ be a compact set, and suppose the set $S=\{z \in K:f(z)=0\}$ is infinite. It thus contains an infinite sequence $(a_n)_{n=0}^\infty$. The Bolzano-Weierstraß gives a convergent subsequence $(a_{n_k})_{k=0}^\infty$ of zeros in $K$, hence in $\Omega$. The identity theorem gives a contradiction.