= Solution
For distinct $x,y\in U$, convexity keeps the segment $x+t(y-x)$ in $U$. Continuity of $Df$ makes
$$
q=\max_{0\leq t\leq1}\|Df(x+t(y-x))-I\|<1.
$$
The <fundamental theorem of calculus along a line segment> gives
$$
\|f(y)-f(x)-(y-x)\|\leq q\|y-x\|,
$$
so $\|f(y)-f(x)\|\geq(1-q)\|y-x\|>0$. Thus \b[$f$ is injective].
The bound $\|Df(a)-I\|<1$ makes $Df(a)$ invertible by the <Neumann series>. The inverse function theorem therefore makes $f$ locally open at every point, so $f(U)$ is open.
Surjectivity need not hold: the identity map on a proper convex open set has proper image. Even for $U=\mathbb R^n$ it can fail. In one dimension, $f(x)=\arctan x$ satisfies $|f'(x)-1|=x^2/(1+x^2)<1$, but its image is $(-\pi/2,\pi/2)$.
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