Let $f$ be a complex-valued holomorphic function defined on an open set $\Omega\subseteq\mathbb{C}$, $f:\Omega\rightarrow\mathbb{C}$, which is not identically zero.
Let $S=\{a\in \Omega :f(a)=0\}$=set of all zeros of $f$.
We know that $S$ is closed in $\mathbb{C}$ as $f$ is continuous. Also, one can easily prove that $S$ is isolated. Therefore, we have
$f\not\equiv0 \implies S$ is isolated.
Hence the contrapositive of the statement is
$S$ is not isolated $\implies f\equiv 0 $
But, $S$ is not isolated implies that is there exists a point of $S$ which is not isolated. But, we also know that $S$ is closed, hence there exists a point of $S$ which is a limit point of $S$.
$S$ has a limit point in $S$ $\implies f\equiv 0$
Question:
But, I know that the above statement is not true. We need the domain $\Omega$ to be connected for the above statement to be true. But I am unable to find any mistake in my argument. Any help will be appreciated.
You can let $\Omega$ be two disjoint open circles and constant with two different constants one $0$ and one $1$ e.g.
In all implications $\Omega$ was assumed to be connected from the start so you cannot drop it anywhere. A srtandard assumption in complex function theory (for holomorphic to be defined etc.) is that the function domain is a "Gebiet" (as it's called in German), i.e. open and connected. Any open set in $\mathbb{C}$ is a disjoint union of these.