What is the proof, that $\sum_{n=0}^{\infty} \frac{1}{2^n}\frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}$ is a metric? What is the proof, that $d(x,y) = \sum_{n=0}^{\infty} \frac{1}{2^n}\frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}$ is a metric? Where $x=(x_1,x_2,\cdots)$ , $y=(y_1,y_2,\cdots)$ and $d_n$ is a metric for $X_n$.
How does one prove the triangle inequality?
https://en.wikipedia.org/wiki/Metric_space#Product_metric_spaces
 A: You can do that component (summand) per component, relying on the fact that each $d_n$ does satisfy the triangle inequality.
In more detail:
$$\begin{align}
\frac{d_n(a,b)}{1+d_n(a,b)}
+ \frac{d_n(b,c)}{1+d_n(b,c)}
&= 
\frac{d_n(a,b)+2d_n(a,b)d_n(b,c)+d_n(b,c)}{(1+d_n(a,b))(1+d_n(b,c))}\\
&= 
\frac{d_n(a,b)+2d_n(a,b)d_n(b,c)+d_n(b,c)}{1+d_n(a,b)+d_n(b,c)+d_n(a,b)d_n(b,c)}
\\&= 
\frac{d_n(a,b)+d_n(b,c)+2u}{1+d_n(a,b)+d_n(b,c)+u}
\\&\geq \frac{d_n(a,b)+d_n(b,c)+u}{1+d_n(a,b)+d_n(b,c)+u}
\\&\geq \frac{d_n(a,b)+d_n(b,c)}{1+d_n(a,b)+d_n(b,c)}
\end{align}$$
setting $u\stackrel{\rm def}{=}d_n(a,b)d_n(b,c)\geq 0$, and using the (easy) fact that if $\beta\geq \alpha > 0$ and $u\geq 0$, then $\frac{\alpha+u}{\beta+u} \geq \frac{\alpha}{\beta}$.
From there, we can rely on the monotonicity of the function $f\colon x\geq 0\mapsto \frac{x}{1+x}$:
$$
\frac{d_n(a,b)+d_n(b,c)}{1+d_n(a,b)+d_n(b,c)} = f(d_n(a,b)+d_n(b,c)) \geq 
f(d_n(a,c)) = \frac{d_n(a,c)}{1+d_n(a,c)}
$$
since by the triangle inequality for $d_n$, we know that $d_n(a,b)+d_n(b,c)\geq d_n(a,c)$.
A: Since $d_i$ is metric, inequality $$d_i(x_i,z_i)\leq d_i(x_i,y_i)+d_i(y_i,z_i)$$ holds in i-th metric space from the product space. 
Also, metrics $d(x,y)$ and  $\frac{d(x,y)}{1+d(x,y)}$ are equivalent. 
(If $d$ is a metric, then $d/(1+d)$ is also a metric). 
Therefore, we have $$\frac{d_i(x_i,z_i)}{1+d_i(x_i,z_i)}\leq\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}+\frac{d_i(y_i,z_i)}{1+d_i(y_i,z_i)},$$
for every index $i$ and
$$\frac{1}{2^i}\frac{d_i(x_i,z_i)}{1+d_i(x_i,z_i)}\leq\frac{1}{2^i}\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}+\frac{1}{2^i}\frac{d_i(y_i,z_i)}{1+d_i(y_i,z_i)},$$
for every $i$. By summing, we get $$d(x,z)\leq d(x,y)+d(y,z).$$
