The title says the question. Here's my proof.
Let $\mathcal{B}_1$ be the basis for $(\mathbb{R}_\text{usual})^2$ and $\mathcal{B}_2$ be the basis for $\mathbb{R}^2_\text{usual}$. The elements of $\mathcal{B}_1$ are of the form $(a, b) \times (c, d)$ which is a rectangle in $\mathbb{R}^2$ while the elements of $\mathcal{B}_2$ are of the form of open balls with radius $r$ given as $$ B_r(x) = \{y \in \mathbb{R}^2 : d(x, y) < r\} $$ We will prove that $\mathbb{R}^2_\text{usual} \subseteq (\mathbb{R}_\text{usual})^2 \subseteq \mathbb{R}^2_\text{usual}$. Onto the second inclusion. Choose $N \in \mathbb{N}$. Split the intervals $(a, b)$ and $(c, d)$ into $N$ equal sub-intervals of length $\epsilon_1$ and $\epsilon_2$ respectively. Let $r = \operatorname{gcd}(\epsilon_1, \epsilon_2)$. This is the radius of our $r-$ball. We wish to fill the rectangle $(a,b) \times (c, d)$ with copies of our $r-$balls. To do this choose $N$ large enough so that $\epsilon_1, \epsilon_2 \to 0$. Now we can approximate the region inside the rectangle $(a, b) \times (c, d)$ with $\mathcal{O}(N^2)$ $r-$ball, meaning that $\mathbb{R}^2_\text{usual}$ refines $(\mathbb{R}_\text{usual})^2$ and that $\mathbb{R}^2_\text{usual} \supseteq (\mathbb{R}_\text{usual})^2$.
Similarly one can show $\mathbb{R}^2_\text{usual} \subseteq (\mathbb{R}_\text{usual})^2$ by using $\epsilon-$rectangles to fill balls of radius $r$. In conclusion, we have $\mathbb{R}^2_\text{usual} \subseteq (\mathbb{R}_\text{usual})^2 \subseteq \mathbb{R}^2_\text{usual}$ which means that $\mathbb{R}^2_\text{usual} = (\mathbb{R}_\text{usual})^2$.
I am not sure about the correctness of my proof. Can someone check my proof. Thanks in advance!!!