Consider $X := (0, 1]$ and the metrics $d_1(x, y) := \left| \frac{1}{x} - \frac{1}{y} \right|$ and $d_2(x, y) = |x - y|$ and show they are topologically equivalent, i.e. for all $x \in X$ $$ \forall \varepsilon > 0 \ \exists r > 0: U_1(x,r) \subset U_2(x,\varepsilon) \quad\text{and} \quad U_2(x,r) \subset U_1(x,\varepsilon), $$ where $U_i(x,r) := \{ y \in X: d_i(x,y) < r\}$ for $i \in \{1,2\}$.
Here's my attempt: Let $x \in X$ and $\varepsilon > 0$ and choose $r_1 := \varepsilon$. For $x,y \in X$ we have $x y \in X$ and therefore \begin{equation*} d_1(x,y) = \frac{|x - y|}{xy} \le d_2(x,y). \end{equation*} Now I have to choose $r_2$ such that $U_1(x,r_2) \subset U_2(x, \varepsilon)$ but I can't find how to do it. I guess it has to depend on $x$ and not only on $\varepsilon$. Choosing $r := \min(r_1, r_2)$ would finish the proof.
Let $C$ be a closed subset of $(X,d_1)$ and $c$ a limit point for the set $C$ with respect to $(X,d_2)$. Then you have that there exists a sequence $c_n\to c$ that lies on $C$ and you have that
$d_1(c,c_n)=|\frac{1}{c}-\frac{1}{c_n}|=d_2(\frac{1}{c},\frac{1}{c_n})=|\frac{1}{c}-\frac{1}{c}|=0$
Because $\frac{1}{x}$ is a continuos function with respect to $(X,d_2)$
So $c$ is a limit point for $C$ with respect $d_1$ but $C$ is $d_1-$ closed so $c\in C$.
This means that $C$ is closed in $d_2$, that means $\tau_1\subseteq \tau_2$.
In a similar way you can prove that $\tau_2\subseteq \tau_1$.
Thus $\tau_1=\tau_2$ so the two metrics are equivalent.