Let $X=\mathbb R\backslash\{0\}$, $d$ be the usual distance on $\mathbb R$ and $d':(x,y)\in X^2\mapsto|x-y|+\left|\frac{1}{x}-\frac1y\right|$.
Are the metric spaces $(X,d')$ and $(X,d)$ uniformly homeomorphic ?
It's quite obvious that both spaces are homeomorphic but I can't seem to find any mapping that is also uniform in both ways. The identity function between these two spaces is only uniformly continuous in one direction. I believe that these two spaces are not uniformly isomorphic but I can't prove it.
If we let $\phi:(X,d')\rightarrow(X,d)$ be a uniform homeomorphism between the two spaces for the sake of contradiction, then we can easily see that $\phi((0,+\infty))=(0,+\infty)$ or $(-\infty,0)$ since $\phi$ is both bijective and uniformly continuous. Since $-\phi$ is also a uniform homeomorphism, then we may suppose WLOG that $\phi((0,+\infty))=(0,+\infty)$. To complete the proof, I want to find a sequence $(x_n)$ of numbers in $(X,d')$ such that $d'(x_n,x_{n-1})\rightarrow0$ but $d(\phi(x_n),\phi(x_{n-1}))\not\rightarrow0$ which would disprove the fact that $\phi$ is uniformly continuous by the sequential caracterization of uniform continuity.