Metrizability of product spaces, Horst Herrlich, Topology I In Horst Herrlich's Topology I, page 117, Statement 4.4.12 the author goes on to prove that product space is metrizable if each component ($\underline X_i$) of the product space is metrizable and the corresp. index set ($I$) is at most countable.
Assuming first the case when $I = N$ is infinitely countable, he then goes on to define such a metric (here each $d_n$ belongs to corresp. $X_n$)
$d((x_n),(y_n)) = max \ \{ \frac 1 n \  d_n(x_n, y_n) | n \in I \}$
If $x = (x_n)$ is element of the product space and $r > 0$, then $\exists m \in N, \forall n \ge m : \frac 1 n < r$
He then goes on to state that it follows:
$S(x,r) = \{ y \in {\displaystyle \prod_{I} X_n} \ | \ d(x,y) < r \ \} = \bigcap \{ p_n^{-1} [S(x_n, n \cdot r)] \ | \ n < m \} \ \ \ [1]$
Here $S(x_n, s) = \{ y \in X_n \ | \ d_n(x_n, y) < s \}$
I really tried to understand but I can't see how [1] just 'follows'. Can someone help in understanding why the equality really holds?
EDIT.
Forgot to mention that  $ diam \ \underline X_i \le 1$
 A: Suppose that $y\in\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}$; then $d_n(x_n,y_n)<nr$ for each $n<m$, so $\frac1n d_n(x_n,y_n)<r$ for each $n<m$. If $n\ge m$, then $\frac1n<r$, so $\frac1n d_n(x_n,y_n)\le\frac1n\cdot1<r$. (You didn’t mention it, but at this point he is clearly assuming that the metrics $d_n$ are bounded by $1$.) Thus, $\frac1n d_n(x_n,y_n)<r$ for all $n\in\Bbb N$, and therefore $$d(x,y)=\max\left\{\frac1n d_n(x_n,y_n):n\in\Bbb N\right\}<r\,,$$ so $y\in S(x,r)$. (The fact that this set has a maximum also depends on the assumption that the metrics $d_n$ have a common upper bound.) This shows that
$$\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}\subseteq S(x,r)\,.$$
Now suppose that $y\in S(x,r)$. Then
$$d(x,y)=\max\left\{\frac1n d_n(x_n,y_n):n\in\Bbb N\right\}<r\,,$$
so $\frac1n d_n(x_n,y_n)<r$ for each $n\in\Bbb N$, and therefore $d_n(x_n,y_n)<nr$ for each $n\in\Bbb N$. In particular, $d_n(x_n,y_n)<nr$ for each $n<m$, so $y_n\in S(x_n,nr)$ for each $n<m$, and therefore $y\in\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}$. Thus,
$$S(x,r)\subseteq\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}\,,$$
so
$$\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}=S(x,r)\,.$$
