Openness w.r.t. these two metrics are equivalent. Suppose $(X,d)$ is a metric space. Define $\delta:X\times X\rightarrow[0,\infty)$, as $$\delta(x,y)=\frac{d(x,y)}{1+d(x,y)}.$$ It is easy to show that $\delta$ is a metric as well, but I am having difficulty in showing that if a subset of $X$ is $d$-open , then it is $\delta$-open too.
Thanks in advance!

Let us take the $d$-open ball $B=\{x\,:\,d(x,x_0)<R\}$ and show that it is $\delta$-open. 
Note that the function $f(x)=\frac{x}{1+x}=\frac{1}{1+\frac{1}{x}}$ is strictly increasing. Therefore $d(x,x_0)<R$ is equivalent to $\delta(x,x_0)=f(d(x,x_0))<f(R)$, i.e.
$$B=\{x\,:\,\delta(x,x_0)<f(R)\}$$
That is, every $d$-ball is also a $\delta$-ball. For the converse notice that $f:(0,\infty)\rightarrow(0,\infty)$ is bijective. Therefore also every $\delta$-ball is a $d$-ball:
$$\{x\,:\,\delta(x,x_0)<R\}=\{x\,:\,d(x,x_0)<f^{-1}(R)\}$$
The topologies generated by $d$ and $\delta$ have the same set of open balls, i.e. they are equal: every $d$-open set is $\delta$-open and vice versa.
