 Solution

For distinct $x,y\in U$, convexity keeps the segment $x+t(y-x)$ in $U$. Continuity of $Df$ makesThe fundamental theorem of calculus along a line segment givesso $\Vert f(y)-f(x)\Vert \ge (1-q)\Vert y-x\Vert >0$. Thus $f$ is injective.

The bound $\Vert Df(a)-I\Vert <1$ 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 $|f^{\prime }(x)-1|=x^2/(1+x^2)<1$, but its image is $(-\pi /2,\pi /2)$.