Consider $\mathbb{R}^{\mathbb N}$ and the two metrics:
$$d(x,y)=\sum^{\infty}2^{-i}\min(1,|x_i-y_i|)$$ and $$d'(x,y)=\sup_{i\in\mathbb{N}}\min(1,|x_i-y_i|).$$
If $O$ is $d$-open, is it $d'$-open? I tried to prove that $d\geq d'$, but I am not sure whether this is the right way to proceed.
In fact you have $d \le d^\prime$, which implies that any open ball $B_{d^\prime}(a,r)$ is included in $B_d(a,r)$.
With that, you can prove that any $d$-open is a $d^\prime$-open.