Showing that product topology (not the one induced by the product metric) is the same as the one induced by the Euclidean metric

Question

My task is to show, that the product topology on $\mathbb{R}^n = \mathbb{R} \times \dots \mathbb{R}$, i.e. the one generated by the basis $$\mathcal{B} := \{U_1 \times \dots \times U_n : U_i \text{ open in } (\mathbb{R},|\cdot|)\}$$ is the same as the one induced by the Euclidean standard metric on $\mathbb{R}^n$.

My idea was to show, that the collection of all open balls in $\mathbb{R}^n$ is also a basis for the product topology, since if two topologies have the same basis, then they are equal. So first of all we have to show that each open ball is in the product topology. So we could use the basis criterion, i.e. if $X$ is a topological space and we have a basis $\mathcal{B}'$ for its topology, then $U$ is open in $X$ if and only if for any $p \in U$ we find $B \in \mathcal{B}'$ such that $p \in B \subseteq U$. Pictorially it is very clear, but formalizing it is not that easy. I just want to know if I am on the right track:

Let $B_r(x)$ be an open ball in $\mathbb{R}^n$ and $p \in B_r(x)$. Then the element $$\left(p_1 - \frac{r - \sqrt{\sum_{k = 1}^n |x_i - p_i|^2}}{2}\right) \times \dots \times \left(p_n - \frac{r - \sqrt{\sum_{k = 1}^n |x_i - p_i|^2}}{2}\right)$$ belongs to the product topology. Now we would have to show that any point of this set lies in the open ball. Somehow, this idea is taken from $\mathbb{R}^2$ and i am not sure if it works in higher dimensions. Any help would be appreciated.

Answer 1

By the brilliant hint of user160738 it is much more elegant to prove first that $$d(x,y) := \sum_{i = 1}^\infty \frac{1}{2^i}d_i(x_i,y_i)$$ is a metric on $\prod_{i = 1}^\infty (M_i,d_i)$, $d_i \leq 1$, which induces the same topology as the product topology (a nice proof can be found in the book general topology by Ryszard Engelking). This can be adapted for norms, i.e. $$N(x) = \sum_{i = 1}^\infty\frac{1}{2^i}N_i(x_i)$$ is a norm on $\prod_{i = 1}^\infty (M_i,N_i)$, $N_i \leq 1$. This is obviously also the case in the finite case, i.e. $\prod_{i = 1}^n (M_i,N_i)$. Now all norms on a finite dimensional space are equivalent, hence induce the same topology. Thanks a lot!