Prove that $f^{-1}$ is uniformly continuous

Question

Suppose $f:X \rightarrow Y$ bijective, $f$ is uniformly continuous, $f^{-1}$ continuous, $X, Y$ are complete. It is required to prove, that $f^{-1}$ is uniformly continuous as well.

My idea was to prove, that $Y$ is compact, therefore, $f^{-1}$ is uniformly continuous on $Y$. But I doubt this idea, because we need to prove, that $Y$ is bounded, which is not the case.

Any suggestions appretiated.

Answer 1

This is false. Let $X=Y=[1,\infty)$ with the usual metric and $f(x)=\sqrt x$. Then $f$ is even Lipschitz but $f^{-1}(x) =x^{2}$ is not uniformly continuous.