= Solution
The <argument principle> states that if $h$ is <meromorphic> on a <neighbourhood> of a positively oriented simple closed <contour> and its interior, with neither zeros nor poles on the <contour>, then
$$
\frac1{2\pi i}\oint_\gamma\frac{h'(z)}{h(z)}\,dz=N-P,
$$
where $N$ and $P$ count its interior zeros and poles with <multiplicity>. This integer is also the <winding number> of $h(\gamma)$ about zero.
Fix $z_0\in U$. Injectivity prevents $f$ from being constant on a <neighbourhood> of $z_0$, so its <Taylor expansion> has the form
$$
f(z)-f(z_0)=(z-z_0)^m h(z),\qquad m\ge1,\quad h(z_0)\ne0.
$$
Choose a small closed disk $D\subset U$, centred at $z_0$, such that $h$ and $mh+(z-z_0)h'$ are nonzero throughout $D$. Then $f-f(z_0)$ has its sole zero at $z_0$ in $D$, of <multiplicity> $m$, while $f'$ has no zero in $D\setminus\{z_0\}$. On its boundary let $\delta=\min|f-f(z_0)|>0$.
Choose $0<|w|<\delta$. For every $0\le t\le1$, the <holomorphic function> $f-f(z_0)-tw$ is nonzero on $\partial D$. By the <argument principle>, its number of zeros is an integer given by a <contour integral> varying continuously with $t$, hence remains $m$. At $t=1$ none of these zeros is $z_0$, so their <derivatives> are nonzero and they are <simple zeros>. Thus there are $m$ distinct points taking the value $f(z_0)+w$. Injectivity forces $m=1$, proving \b[$f'(z_0)\ne0$ everywhere in $U$].
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