Define $f$ to be complex analytic on some subset $S$ of $\mathbb{C}$ if $f$ is given locally by a power series (I'm not requiring $S$ itself to be open.)
If $S$ is connected and open, then if there is a sequence of distinct numbers in $f^{-1}(\{0\})$ with a limit point in $S$, then $f\equiv 0$ on $S$. The proof I saw of this went something like: Show that at the limit point $z_0$, the power series for $f$ must have zero coefficients, and so $f$ vanishes in some open disk around $z_0$. Thus $\text{interior}[f^{-1}(\{0\})]$ is nonempty, open and closed in $S$, hence equal to $S$ by connectedness. Does the proof require $S$ to be open?
Also I'm wondering if an almost-converse is true: If $S$ contains a limit point, and every complex analytic function on $S$ with a limit point of zeroes in $S$ is identically zero, then is $S$ connected?