= Solution
Let $f$ be <holomorphic> on a neighborhood of the closure of a bounded region with piecewise smooth positively oriented boundary, with no boundary zeros and with $f$ not identically zero. The <argument principle> states
$$
\boxed{\frac1{2\pi i}\int_{\partial G}\frac{f'(z)}{f(z)}\,dz=\sum_{a\in G:f(a)=0}m_a,}
$$
where $m_a$ is the zero <multiplicity>. More generally a closed contour avoiding the zeros gives $\sum_a m_a\operatorname{Ind}(\Gamma,a)$, and the integral is the <winding number> of $f\circ\Gamma$ about zero.
To prove it, factor near a zero $a$ as $f(z)=(z-a)^{m_a}h(z)$ with $h(a)\ne0$. Then $f'/f=m_a/(z-a)+h'/h$. Remove small disjoint circles about the finitely many zeros in the region. On the remainder $f'/f$ is <holomorphic>, so the <Cauchy integral theorem> reduces the boundary integral to the sum of the small-circle integrals. Each is $2\pi i m_a$ because $h'/h$ is <holomorphic> on its small disc. This proves the formula. Along a contour, the imaginary part of $d\log f$ is the change in a continuous argument, giving the winding interpretation and the indexed version.
Now choose a closed disc $\overline{D(z_0,r)}\subset\Omega$ on which $z_0$ is the only zero of the nonconstant limit $f$. Its boundary has $\min|f|>0$. <Local uniform convergence> gives $\max|f_n-f|<\min|f|$ there for all sufficiently large $n$. The contour functions $f+t(f_n-f)$, $0\le t\le1$, have no boundary zero. Their <winding numbers> are constant under this homotopy, so the <argument principle> shows that $f_n$ has exactly the same positive number $m$ of zeros, counted with <multiplicity>, inside the disc. This is the <Rouche theorem> argument, here derived from the contour count.
Choose a zero $z_n$ in that disc. For every $0<\epsilon<r$, the compact annulus $\epsilon\le|z-z_0|\le r$ contains no zero of $f$, so <uniform convergence> makes $f_n$ nonzero on it eventually. Therefore all of the chosen zeros eventually lie within $\epsilon$ of $z_0$. Consequently
$$
\boxed{f_n(z_n)=0,\qquad z_n\longrightarrow z_0.}
$$
The maps $f_n$ are eventually not identically zero, since they converge at any fixed point where $f$ is nonzero; thus their zero counts above are legitimate.
A double zero need not persist as a double zero. On the <unit disc> take $f(z)=z^2$, $f_n(z)=z^2+1/n$. Its zeros $\pm i/\sqrt n$ converge to zero but are simple, while $f_n'$ vanishes only at zero and $f_n(0)=1/n\ne0$. Hence \b[the additional simultaneous <derivative> condition cannot always be imposed]. <Local uniform convergence> preserves total nearby <multiplicity>, not its concentration at a single point.
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