If $d(x,y)$ is a metric on $X$, then $d'(x,y)=\frac{d(x,y)}{1 + d(x,y)}$ and $d(x,y)$ generates the same topology. If $d(x,y)$  and $d'(x,y)$ are a metrics on $X$, then $d'(x,y)=\frac{d(x,y)}{1 + d(x,y)}$ and $d(x,y)$ generates the same topology.
My attempt:-
 Let $\mathscr T$ be the topology generated by $d$ and $\mathscr T'$ be the topology generated by $d'$. Let $U\in \mathscr T\implies U=\bigcup_{\alpha\in \Lambda}B_d(x_\alpha, r_\alpha).$ So, $x\in B_d(x_\alpha, r_\alpha)$. We want to prove there is a $B_{d'}(x,\delta)\subset B_d(x_\alpha, r_\alpha).$ If $r=\min\{.9,r_{\alpha}\}. So,B_{d}(x_\alpha,r)\subset B_d(x_\alpha, r_\alpha). $ I could prove that $B_{d'}(x_\alpha,\frac{r}{1-r})\subset B_{d}(x_\alpha, r)$. Let $y\in  B_d(x_\alpha, r_\alpha)$, let $\epsilon=\min\{r_\alpha-d(x_\alpha,y),.9\}.$ Then, there is an $B_{d'}(y,\frac{\epsilon}{1-\epsilon})\subset B_{d}(y, \epsilon)\subset B_{d}(x_\alpha, r_\alpha).$ Hence, I can write $U$ as the union of open balls of $d'$. Hence $U\in \mathscr T'$. If my proof is correct, I can write the converse similarly. By using $B_d(x,\frac{r}{r+1})\subset B_{d'}(x,r)$. Please help me to rectify my errors.
 A: It is enough to show (and necessary too) that every $d$-ball around a point contains a $d'$-ball around that point and vice versa. In that case both metrics generate the same topology: if $O$ is $d$-open, and $x \in O$ then for some $r>0$ $B_d(x,r) \subseteq O$ and then we can find $r'>0$ such that $B_{d'}(x,r') \subseteq B(x,r)$ (any $d$-ball around $x$ contains a $d'$ ball around $x$) and then $x$ is a $d'$-interior point of $O$, and we can do this for all $x$, so $O$ is $d'$-open too. Interchanging the rôles of $d'$ and $d$ we get the reverse too.
Now, the main fact I'll use to show this mutual ball inclusions is that $d' = f \circ d$ where $f: \mathbb{R}_0^+ \to \mathbb{R}_0^+$ is the function $f(x) = \frac{x}{1+x}$, with $f(0) = 0$ and $f$ continuous at $0$ (in the usual topology on $\mathbb{R}_0^+$), and also that $x < f(x)$ for all $x$.
Let $x \in X$ and $r>0$. Then by continuity of $f$ at $0$ we find some $r'>0$ such that $0 \le t < r'$ implies $f(t) < r$ and this means that $B_{d}(x,r') \subseteq B_{d'}(x,r)$: if $y$ obeys $d(x,y) < r'$ then $d'(x,y) = f(d(x,y)) < r$. This shows one ball-inclusion we need.
If $x \in X$ and $r>0$, then $B_{d'}(x,r) \subseteq B_d(x,r)$ because if $d'(x,y) < r$ then $d(x,y)) < f(d(x,y)) = d'(x,y) < r$ as well). This shows the ball inclusion in the other direction.
Your $0.9$ term seems a bit random (maybe inspired on the proof that the truncated metric $d''(x,y) = \min(d(x,y), 1)$ also induces the same topology as $d$?). If we abstract away the exact formula of $d'$ by reasoning a little on the $f$, the proof becomes easier. Also, use the mutual inclusion criterion to show equivalence of metrics. It was what you were trying to do as well, I think. 
A: i) $d(x_0,y)<\varepsilon$ so that $$ d'(x_0,y)\leq
d(x_0,y)<\varepsilon $$
Hence $B^d(x_0,\varepsilon)\subset B^{d'}(x_0,\varepsilon)$
ii) If $d'(x_0,y)< \varepsilon < 1/2$, then $$
d(x_0,y)< \varepsilon +\varepsilon d(x_0,y) \Rightarrow d(x_0,y)<
\frac{\varepsilon }{1-\varepsilon } < 2\varepsilon $$
$B^{d'}(x_0,\varepsilon)\subset B^{d}(x_0,2\varepsilon)$
