= Solution
Uniform convergence on compact subsets makes $f$ continuous. For every triangle whose closure lies in $U$, uniform convergence on its boundary permits passage through the contour integral:
$$
\int_{\partial\Delta}f(z),dz
=\lim_{n\to\infty}\int_{\partial\Delta}f_n(z),dz=0.
$$
<Morera's theorem> implies that $f$ is holomorphic. If a compact set $K\Subset U$ is surrounded by a finite union of contours at positive distance from $K$, the <Cauchy derivative formula>[Cauchy integral formula for derivatives] gives
$$
f_n'(z)-f'(z)
=\frac1{2\pi i}\int_\Gamma
\frac{f_n(\zeta)-f(\zeta)}{(\zeta-z)^2},d\zeta.
$$
The uniform bound on the surrounding compact set proves that $f_n'\to f'$ uniformly on $K$.
If $f$ had distinct zeros $c$ and $d$, choose disjoint small closed discs around them whose boundary contains no zero of $f$. Uniform convergence and <Rouché's theorem> imply that, for large $n$, $f_n$ has a zero in each disc, contradicting uniqueness of $c_n$. Thus $f$ has at most one zero.
For an example on the unit disc, take
$$
f_n(z)=z-(1-1/n),
\qquad c_n=1-1/n.
$$
Then $f_n\to f(z)=z-1$, which has no zero in the open disc. In general, <Hurwitz's theorem> shows that $f$ is zero-free exactly when the unique zeros escape every compact subset of $U$:
$$
\boxed{
\text{for every compact }K\Subset U,
\quad c_n\notin K\text{ eventually}.}
$$
Indeed, an interior accumulation point of $c_n$ is a zero of $f$, while a zero of $f$ forces the unique $c_n$ into each of its sufficiently small neighbourhoods.
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