Show this metric generates the product topology on $X$ Let $(X_n, d_n)$ be a sequence of metric spaces. Show that the function $ d: X \times X \to \mathbb R^+$ on the product space $X: = \prod_n X_n$ defined by 
$$d ((x_n)_{n = 1}^\infty, (y_n)_{n=1}^\infty ) := \sum_{ n=1}^\infty 2^{-n} \frac{ d_n(x_n,y_n)} { 1+ d_n (x_n,y_n) } $$
is a metric on $X$ which generates the product topology on $X$.

I showed that $d$ is actually a metric, which was easy. To show that this metric generates the product topology I think I need to show: 
(i) Each ball $ B((x_n)_{n=1}^\infty  , \epsilon )$ is open in the product topology. 
(ii) For any $B(x_n , \epsilon) \subset X_n$, $\pi_n ^{-1} (B(x_n , \epsilon)) \subset X$ is the union of finite intersections of balls in $(X, d)$.
But I couldn't even get started to do (i). Any help is appreciated.
 A: Note that if $(X,d)$ is metric space, then $d'=d/(1+d)$ generates same topology of $(X,d)$. So we only prove this proposition: 

Let $(X_n,d_n)$ be a sequence of metric spaces, and $d_n(x,y)\le 1$ for all $n$ and $x,y\in X_n$, then $d((x_n),(y_n))=\sum_n 2^{-n} d_n(x_n,y_n)$ generates the product topology of $X=\prod_n X_n$.

At first, we prove that for each $a=(a_n)_{n=1}^\infty\in X$ and $r>0$, there is a open basis $V$ of $X$ satisfy that $a\in V\subset B_d(a,r)$.
Let $N$ be a natural number that satisfy $2^{-N}\le r/2$. Consider
$$
V= B_1 (a_1,r/2)\times \cdots\times B_N (a_N,r/2)\times X_{N+1}\times X_{N+2}\times\cdots
$$
(where $B_i(x,r)$ is a open ball in $X_i$.) If $x\in V$ then $d(a_i,x_i)<r/2$ for 
$i=1,2,\cdots, N$. Therefore
$$
\begin{aligned}
d(a,x)=\sum_{n=1}^\infty 2^{-n}d_n(a_n,x_n)&\le \frac{1}{2}\sum_{n=1}^N 2^{-n}r + \sum_{n>N}2^{-n}\\
&=(1-2^{-N})\cdot\frac{r}{2}+2^{-N}\\
&< \frac{r}{2}+\frac{r}{2}=r
\end{aligned}
$$
so $x\in B_d(a,r)$. 
Finally, you prove that for each $a\in X$ and for each basis $V$ of $X$ that contaning $a$ 
there is $r$ satisfy that $a\in B_d(a,r)\subset V$. It is easy to prove so I leave the proof of this part for yours.
