$|\nabla f (x)| =1$ implies $f$ linear?

Yes, this is true: smooth global solutions of the eikonal equation are affine. The proof is similar to Solution of eikonal equation is locally the distance from a hypersurface, up to a constant. It goes like this:

1. By the mean value theorem, $f$ is $1$-Lipschitz.
2. Trajectories of steepest ascent/descent, i.e., solutions of the ODE $x'(t)=\nabla f(x(t))$ both forward and backward in time, are straight lines. Indeed, along such a curve we have $$|x(t)-x(s)| \ge |f(x(t))-f(x(s))|=|t-s|$$ which, the curve being unit speed, implies that it's a line.
3. Let $\Gamma$ be any level surface of $f$ (WLOG, $G=\{f=0\}$), and $L$ a normal line to this set. By item 2 above, $L$ a path of steepest ascent/descent. Choose coordinates so that $L$ is the $x_1$-axis. The restriction of $f$ to $L$ is either $x_1$ or $-x_1$. The $1$-Lipschitz condition implies $\operatorname{dist} (te_1,\Gamma)=|t|$ for all $t\in\mathbb{R}$. Hence, $\Gamma=\{x:x_1=0\}$.
4. Since all level surfaces are planes, and they are disjoint, they must be parallel. After rotation, $f$ is a function of $x_1$ only: more specifically, either $x_1$ or $-x_1$.