 Solution

For arbitrary $y\in {\mathbb{R}}^2$, the continuous map $g_y(x)=g(x)-y$ satisfies $\Vert g_y(x)-x\Vert \le K+\Vert y\Vert$. Applying the preceding Brouwer fixed-point theorem argument with this positive bound gives $g_y(t)=0$. Hence $g(t)=y$, and $g$ is surjective.