Best Answer:
Well, this is certainly true if f is not identically zero. You, of course, forgot to add this assumption to your assertion. So, I'll suppose that f is not identically 0. If this is the case, your assertion follows from the following theorem, whose proof you find in any book on Complex Analysis:

Let V and f be as in your assertion and let Z(V) be the set of all zeroes of f in V. If Z(V) has a limit point in V, then f is identically 0 in V (this follows from the fact the holomorphic functions are given by power series).

Suppose K ⊂ V is compact. Then (Heine/Borel theorem), K is closed and bounded. So, Z(K) = {z ∈ K : f(z) = 0} is bounded. If Z(K) is infinite, then (Bolzano/Weierstrass theorem) Z(K) has a limit point a. Since Z(K) ⊂ Z(V) and Z(K) ⊂ K, a is automatically a limit point of both Z(V) and K. Since K ⊂ V is closed, a ∈ K and, therefore, a ∈ V. We conclude Z(V) has a limit point in V (the domain of f). According to the theorem we mentioned, f, contrarily to our assumption, is identically 0. It follows that, in every compact subset of V, f must have finitely many (possibly none) zeroes.

The assumption that f is not identically 0 is, of course, essential for your assertion to be always true for every connected and open subset of C.

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