I have multiple ideas of how to try show this, I want to know if the approaches are correct and help correct any misunderstanding in my arguments below:
We know $B_1 = \{ \text{ all open balls in } \mathbb{R}^2 \}$ is a basis for the standard topology on $\mathbb{R}^2$ and
$B_2$ = $\{U \times V: \text{ U and V are open in } \mathbb{R} \}$ is a basis on the product topology on $\mathbb{R}^2$. If we show $B_1 = B_2$ would we be done? But clearly equality does not hold since a open ball cannot be written as a product of two sets in $\mathbb{R}$. But what if we showed $B_1$ and $B_2$ generate each other, that is a open ball can be written as union of open rectangles and vice versa, would that work?
Another approach could be to define $\mathscr T_1$ = $\{U: U \text{ open in } \mathbb{R}^2\}$ (standard topology) and
$\mathscr T_2$ = $\{A: A \text{ is a union of sets in } B_2\}$ then try show $\mathscr T_1$ = $\mathscr T_2$. We could do this by saying any set in $\mathscr T_2$ will be open in $\mathbb{R}^2$ so will be in $\mathscr T_1$. And since the set of all open rectangles is a basis for $\mathbb{R}^2$ any open ball can be written as union of sets in $B_2$ hence is in $\mathscr T_2$.
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