This has nothing to do with convexity. The exact formulation is this:
Let $\Omega$ be an arbitrary region, and $f_n\to f$ is a sequence of holomorphic functions converging uniformly on compacts in $\Omega$.
If $f\neq 0$ (this is an important condition!), and $D$ is a region,
such that $\overline{D}\subset \Omega$, and $f(z)\neq 0$ on $\partial D_1$, then there exists $N$ (dependng on $D$) such that for $n\geq N$, $f$ and $f_n$ have the same number of zeros
in $D$.
For the proof, you find an intermediate region $D_1$ such that
$D\subset D_1\subset\overline{D_1}\subset\Omega$, and $\partial D_1$ is piecewise smooth, and the number of zeros of $f$ in $\overline{D_1}$ is the same as the number of zeros in $D$. Such region exists since $\overline{D}\subset\Omega$, and since the zeros
of $f$ are isolated and $f(z)\neq 0,\; z\in\partial D$.
Then the number
of zeros $f$ in $D$ is the same as in $D_1$ and is equal to
$$\frac{1}{2\pi i}\int_{\partial D_1} \frac{df}{f},$$
and the number of zeros of $f_n$ in $D_1$ is
$$\frac{1}{2\pi i}\int_{\partial D_1} \frac{df_n}{f_n}.$$
Since $f_n\to f$ uniformly on the compact set $K:=\overline{D_1}\backslash D$, and $f(z)\neq 0$ on $K$, we conclude that for $n\geq N$, $f_n$ has the same number of zeros
in $D$ and $D_1$. Since the integrals for $f_n^\prime/f_n$ converge to the
integral for $f'/f$, and all these integrals are integers, we conclude that they are equal for $n\geq N$.
Refs. This is stated in full generality in the book
A. I. Markushevich, Theory of functions of a complex variable,
vol. I, Ch. IV, Sect 3 (p. 426 of the Russian original).
Special cases are in:
Ahlfors, Complex Analysis, p. 178, Theorem 2,
Titchmarsh, The theory of functions, sect. 3.45,
Marshall, Complex Analysis, Theorem 8.8,
But all these special cases have the same proof as the general theorem whose complete proof I wrote.
