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.

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.

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 $$ d(x_0,y)< \varepsilon +\varepsilon d(x_0,y) \Rightarrow d(x_0,y)< \frac{\varepsilon }{1-\varepsilon } < 2\varepsilon $$

$B^{d'}(x_0,\varepsilon)\subset B^{d}(x_0,2\varepsilon)$