Let $\bar d(a,b)=\min\{|a-b|,1\}$ the standard bounded metric on $\mathbb R$. Define $$D=\sup\left\{\frac{\bar d(x_i,y_i)}{i}\right\}.$$
I proved it's indeed a metric on $\mathbb R^{\mathbb N}$ but in the Munkres they say that it induce the product topology, but I don't understand why. I mean, why for all $$V=\prod_{i=1}^\infty U_i$$ open where $U_i=\mathbb R$ but finitely many $i$, then $V=B_D(x,\varepsilon)$ for an $x\in \mathbb R^{\mathbb N}$ and an $\varepsilon>0$ ?
Let us fix some notation first. I will denote by $\tau_d$ the topology on $\mathbb{R}^\mathbb{N}$ generated by the metric $D$, and by $\tau_p$ the product topology on $\mathbb{R}^\mathbb{N}$. The first topology has as a basis the set of open balls, and the second one has as a basis the sets given by the product of open subsets of $\mathbb{R}$, with only finitely many of the factors being proper subsets.
Your question at the end is if these basis are the same. The answer is no, and this is not a problem. We want to show that $\tau_d = \tau_p$, so we need to show that an element of $\tau_d$ also belongs to $\tau_p$, and vice-versa. This is not related to equality between bases. For example, the euclidean metric and the taxicab metric both induce the same topology on $\mathbb{R}^k$, but an open ball in the euclidean metric cannot be given by an open ball in the taxicab metric.
Let us then show that $\tau_d=\tau_p$. We will show first that $\tau_p \subseteq \tau_d$. The product topology on $\mathbb{R}^\mathbb{N}$ has as subbasis the sets of the form $\pi_j^{-1}(U)$, where $U \subseteq \mathbb{R}$ is open and $\pi_j:\mathbb{R}^\mathbb{N} \to \mathbb{R}$ is the projection onto the $j$-th factor. To show that $\tau_p \subseteq \tau_d$, it is sufficient to show that every subbasic element of $\tau_p$ belongs to $\tau_d$. So, if $x=(x_i)_{i \in \mathbb{N}} \in \pi_j^{-1}(U)$, then $x_j \in U$ and so there exists an $r>0$ such that $B(x_j,r) \subseteq U$. Notice that, by replacing $r$ by a smaller one if necessary, we may assume that $r<1$. Now we claim that $B_D(x,r/j) \subseteq \pi_j^{-1}(U)$. If $y \in B_D(x,r/j)$, then $D(x,y)<r/j$ and by the definition of $D$ we conclude that $\overline{d}(x_j,y_j)/j < r/j$, so $\overline{d}(x_j,y_j) < r$. By the choice of $r$ we made before, $\overline{d}(x_j,y_j) = \lvert x_j-y_j \rvert$, so we conclude that $y_j \in B(x_j,r) \subset U$, i.e., $y \in \pi_j^{-1}(U)$. It follows that $B_D(x,r/j) \subset \pi_j^{-1}(U)$, so $\pi_j^{-1}(U)$ is also open in the topology given by $D$.
For the converse, consider a ball $B_D(x,r) \subset \mathbb{R}^\mathbb{N}$ and fix $r'<r$. Notice that if $i_0$ is the smallest index satisfying $r'i_0 \geq 1$, then we have $\overline{d}(x_i,y_i) \leq r'i$ trivially for $i \geq i_0$. So, intuitively, being in the ball $B_D(x,r)$ imposes conditions on the coordinates $y_i$ only for $i<i_0$. Motivated by this, consider a ball $B(y_i,\varepsilon_i) \subset \mathbb{R}$ for each $i<i_0$, with $\varepsilon_i=r'i-\overline{d}(x_i,y_i)$. We claim that $$ U := B(x_1,\varepsilon_1)\times \dotsm \times B(x_{i_0-1},\varepsilon_{i_0-1}) \times \mathbb{R} \times \mathbb{R} \times \dotsm \subset B_D(x,r). $$
Indeed, if $z \in U$, by the observations we made above, we have $\overline{d}(z_i,x_i)<r'i$ for $i \geq i_0$. For $i<i_0$, by the choice of $\varepsilon_i$, we have $$ \overline{d}(z_i,x_i) \leq \overline{d}(z_i,y_i)+\overline{d}(y_i,x_i) \leq r'i-\overline{d}(x_i,y_i)+\overline{d}(x_i,y_i) = r'i. $$
From the calculation above we conclude that $$ \frac{\overline{d}(z_i,x_i)}{i} \leq r' \ \forall i \in \mathbb{N} \implies D(z,x) = \sup\left\{\frac{\overline{d}(z_i,x_i)}{i}\right\} \leq r' < r \implies z \in B_D(x,r). $$
This shows that $U \subset B_D(x,r)$, and since $U$ is open in $\tau_p$, the ball $B_D(x,r)$ is also open in this topology. Since the balls form a basis for $\tau_d$, we conclude that $\tau_d \subseteq \tau_p$.
