$\|h(x)-h(y)\|\geq 3\|x-y\|$ prove that image of open is open
Solution 1:
Note that $h$ is injective (if $h(x)=h(y)$, then $x=y$) hence there is an inverse $g$ and $\|g(x)-g(y)\| \le {1 \over 3} \|x-y\|$, so $g$ is continuous.
In particular, if $U$ is open, then $g^{-1}(U) = h(U)$ is open.