Let me use a very similar path path which can be adapted in you case.
Lemma : Let (X,d) be metric then $\tilde{d}(x,y) = \min\{d(x,y),1\}$
is a distance topologically equivalent to $d$.
Using this Lemma we can prove :
Theorem : $d(x,y) := \sum\limits_{i=0}^{\infty} 2^{-i}\tilde{d}(x_i,y_i)$ induces
the product topology.
Let $\tau_d$ the topology induced by this metric. We have to show that $\tau_X = \tau_d$. If we denote with $\pi_i : X \longrightarrow X_i$ the projection, then $d_i(\pi_i(x),\pi_i(y)) = d_i(x_i,y_i) = 2^i(2^{-1}d_i(x_i,y_i)) \leq 2^id(x,y)$ hence $\pi_i$ is $2^i$-lipschitz hence every projection is continuos with respect to $\tau_d \Rightarrow \tau_X < \tau_d$.
To see the converse it is sufficient to show that every ball of $\tau_d$ is open in $\tau_X$. Let $B = B_d(x,\epsilon) \subseteq X$ and $y \in B$. Then it exists $\delta > 0 : B_d(y,\delta) \subseteq B$.
We have to show that exists $U$ open in $\tau_X$ such that $y \in U \subseteq B_d(y,\delta)$.
Let $n_0 \in \mathbb{N} : \sum\limits_{i = n_0 + 1}^{\infty} 2^{-i} < \frac{\delta}{2}$ (exists by convergence of the series).
We set $U := \bigcap\limits_{i=0}^{n_0} \pi_{i}^{-1}(B_{d_i}(y_i,\frac{\delta}{4})) = B_{d_0}(y_0,\frac{\delta}{4}) \times \cdots \times X_{n_0+1} \times \cdots$.
If $z \in U$ with $d_i(y_i,z_i) < \frac{\delta}{4} \forall i \leq n_0$ then
$$d(y,z) = \sum\limits_{i=0}^{\infty} 2^{-i}d(x_i,y_i) = \sum\limits_{i=0}^{n_0}2^{-i}d(x_i,y_i) + \sum\limits_{i=n_0+1}^{\infty} 2^{-i}d(x_i,y_i) \leq \frac{\delta}{4}\sum\limits_{i=0}^{n_0}2^{-i} + \frac{\delta}{2} < \delta$$
So $U \subseteq B_d(y,\delta)$.
