I am encountering the same statement as in the question: Equivalent Conditions on Equivalent Metrics:
Show that two metrics $\rho$ and $\rho'$ are equivalent if there exist $\alpha,\beta,M>0$ such that $\rho(x,y)\leq M$ implies $\alpha\rho(x,y)\leq\rho'(x,y)\leq\beta\rho(x,y)$.
Here are some sketch of my proof to this statement:
Proof. For every $x\in X$, let $B_\rho(x,r)$ be the open ball centered at $x$ of radius $r>0$ under the metric $\rho$. Set $$r'=\min\{r,M\}.$$Then we have $r'\leq r$ and hence $B_\rho(x,r')\subseteq B_\rho(x,r)$. Note that for every $y\in B_\rho(x,r')$ we have $\rho(x,y)<r'\leq M$. Consequently, $$B_{\rho'}(x,\alpha r')\subseteq B_\rho(x,r')\subseteq B_\rho(x,r)$$ as for every $y\in B_{\rho'}(x,\alpha r')$, $$\rho(x,y)\leq\frac{1}{\alpha}\rho'(x,y)<\frac{\alpha r'}{\alpha}=r'.$$
Similarly, let $B_{\rho'}(x,r)$ be the open ball centered at $x$ of radius $r>0$ under the metric $\rho'$. Set $$r''=\min\{r,\beta M\}.$$ Apparently, $r''\leq r$ and hence $B_{\rho'}(x,r'')\subseteq B_{\rho'}(x,r)$. Moreover, we have $$B_\rho(x,r''/\beta)\subseteq B_{\rho'}(x,r'')\subseteq B_{\rho'}(x,r)$$ since for every $z\in B_\rho(x,r''/\beta)$, we have $\rho(x,z)<r''/\beta\leq M$ and by assumption $$\rho'(x,z)\leq\beta\rho(x,z)<\beta\cdot\frac{r''}{\beta}=r''.$$
Can anyone help to check if my proof is fine? Thanks.
Edit. I have found a flaw in the first part of the proof. The inequality $\rho(x,y)\leq\rho'(x,y)/a$ is true if $\rho(x,y)\leq M$. However, what we want to show is exactly $\rho(x,y)\leq r'\leq M$. Thus, the first part of the proof is invalid, but I can't find any way to fix it.