There is an exercise 9.3#1 from "Topology without tears":
Let $(X_i, d_i), i\in \mathbb{N}$, countable infinity of metric spaces, where every metric is bounded: $\forall X_i\forall a, b \in X_i, d_i(a, b) \le 1$
Let $D_1, D_2=(\prod_{n=1}^\infty(X_i)\times\prod_{n=1}^\infty(X_i)) \to \mathbb{R}$, two metrics on product space
$$D_1\left(\prod_{n=1}^\infty(a_i),\prod_{n=1}^\infty(b_i)\right)=\sup \{d_i(a_i, b_i): i \in \mathbb{N}\} $$
$$D_2\left(\prod_{n=1}^\infty(a_i),\prod_{n=1}^\infty(b_i)\right)=\sum_{n=1}^\infty\frac{d_i(a_i, b_i)}{2^i} $$
We need to show their equivalence (equivalence of topologies induced by this metrics)
But how could this metrics induce the same topologies? Second one induce usual countable product topology, with base $O_1\times O_2 \times ... \times O_n\times X_{n+1}\times X_{n+2} \times ...$ (where $O_i$ is any set from the base of $X_i$)
and first should induce box topology, $O_1\times O_2 \times ... \times O_n\times O_{n+1}\times O_{n+2} \times ...$ (this is not correct, I think, but I don't understand why — i.e. if we imagine open ball of radius 0.5, it will be a product of open balls of radius 0.5 on every $X_i$, and this ball will be always a subset of $X_i$, not equal to $X_i$)
So, what i miss?
I believe this exercise is wrong. Your remark about the open ball of radius $1/2$ being the products of open balls of radius $1/2$ is also wrong in general [it fails for $X_i=[0,1],\,\forall i$ and the usual Euclidean metric because $(\frac{1}{2}-2^{-i})_{i\in\Bbb N}$ is in the product of the open balls about zero but $\sup_i d(a_i,b_i)=1/2$ in this instance, not $<1/2$).
It is hopefully clear that the topology induced by $D_2$ is coarser than the topology induced by $D_1$. However, the converse inclusion is false. Your idea about the box topology works when there isn't a "continuum" of values attained by the metric, whatever that means; your idea about the box topology works in the counterexample below.
Take $X_i$ to be the discrete metric space on two points, for every $i\in\Bbb N$. The metric of $D_1$ on $\prod_i X_i=\{0,1\}^{\Bbb N}$ is again just the discrete metric; $d(a,b)=\begin{cases}1&a\neq b\\0&a=b\end{cases}$. This cannot be topologically equivalent to the product metric since the point $\{(0,0,0,\cdots)\}$ is open with respect to $D_1$ but is not open in the product topology.