Here was my effort that the product topology is the same as the euclidean topology on $\mathbb{R}^n$, is it ok?
The product topology on $\mathbb{R}^n$ is the same as the standard topology on $\mathbb{R}^n$. Let $U$ be an open set in $\mathbb{R}^n$. Let $p \in U$. Then there is an $\epsilon>0$ such that $B_{\epsilon}(p)=\{(x_1,…,x_n) \in \mathbb{R}^n|\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+\cdots+(x_n-y_n)^2}<\epsilon\} \subset U$. Then $(x_1-\frac{\epsilon}{\sqrt{n}},x_1+\frac{\epsilon}{\sqrt{n}}) \times (x_2-\frac{\epsilon}{\sqrt{n}},x_2+\frac{\epsilon}{\sqrt{n}})\times \cdots \times (x_n-\frac{\epsilon}{\sqrt{n}},x_n+\frac{\epsilon}{\sqrt{n}}) \subset B_{\epsilon}(p)=\{(x_1,…,x_n) \in \mathbb{R}^n|\sqrt{(x_1-y_1)^2+ \cdots +(x_n-y_n)^2}<\epsilon \} \subset U$.
So $U$ is open in $\mathbb{R}^n$.
Conversely suppose that $U$ is open in $\mathbb{R}^n$. Then for each point $(x_1,…,x_n) \in U$ there is a basis element $(u_1,v_n) \times (u_2,v_2) \times \cdots \times (u_n,v_n)$ centered about $(x_1,…,x_n)$ such that $(u_1,v_1) \times (u_2,v_2) \times \cdots \times (u_n,v_n) \subset U$. Then $(x_1,…,x_n)=(\frac{u_1+v_1}{2},…,\frac{u_n+v_n}{2})$. Choose $\epsilon=\text{min}\{\frac{|u_1-v_1|}{2},…,\frac{|u_n-v_n|}{2}\}$. Then $B_{\epsilon}((x_1,…,x_n))\subset (u_1,v_1) \times (u_2,v_2) \times \cdots \times (u_n,v_n) \subset U$. So $U$ is open in $\mathbb{R}^n$.
Also I realize the display is messed up on my post, and I do not know how to fix it.
Best Answer
Since you have taken the coordinates of $p$ as $(y_1,y_2,...,y_n)$ you probably wanted to take the box $\prod_{i=1}^n(y_i-\epsilon/\sqrt n,y_i+\epsilon/\sqrt n)$. You are also just writing $U$ is open in $\Bbb R^n$ without stating which topology.
Note that $x_i\in(y_i-\epsilon/\sqrt n,y_i+\epsilon/\sqrt n)\implies(x_i-y_i)^2<\epsilon^2/n\implies\sqrt{\sum_{i=1}^n(x_i-y_i)^2}<\sqrt{n\cdot\epsilon^2/n}=\epsilon$ so this box is indeed a subset of $B_\epsilon(p)\subseteq U$.
For the converse, not that $y\in B_\epsilon(x)\implies|y_i-x_i|<\epsilon\implies y_i\in(x_i-\epsilon,x_i+\epsilon)\subseteq (x_i-|u_i-v_i|/2,x_i+|u_i-v_i|/2$ and hence $B_\epsilon(x)\subseteq\prod_{i=1}^n(u_i,v_i)$.