Would the following be correct?:
Let $(X_1\times X_2, T_1\times T_2)$ be the product topology of $(X_1,T_1)$ & $(X_2,T_2)$, defined as the least fine topology that makes the projections $$p_1:X_1\times X_2\rightarrow X_1: (x_1, x_2)\rightarrow x_1$$ & $$p_2:X_1\times X_2\rightarrow X_2:(x_1,x_2)\rightarrow x_2$$ continuous.
By a previous lemma, $$T_1\times T_2=\{ p_1^{-1}(U_1)\cap p^{-1}_2(U_2): U_1\in T_1, U_2\in T_2\} .$$ Now, $p^{-1}_1(U_1)=U_1\times X_2$ and $p^{-1}_2(U_2)=X_1\times U_2$ implies $p_1^{-1}(U_1)\cap p_2^{-1}(U_2)=U_1\times U_2$ and thus $$T_1\times T_2=\{ U_1\times U_2:U_1\in T_1, U_2\in T_2\} .$$
The reason I find this confusing is that I do not see a mistake in my reasoning, yet the author of my book merely claims that any element $U\in T_1\times T_2$ is of the form $$\bigcup _{\alpha ,\beta} \{U^{\alpha}_1\times U^{\beta}_2\}$$ for $U^{\alpha}_1\in T_1$, $U^{\beta}_2\in T_2$, which doesn't contradict my reasoning, but if I was correct then certainly the author would mention the result I've shown, yet he doesn't.
I would appreciate any help.
The correct statement is the second one: an open set is a union of some family of open rectangles $U_i \times V_i$ with $U_i$ open in $X_1$ and $V_i$ open in $X_2$ for all $i$.
The first part shows that in order to make projections continuous (and because topologies are closed under finite intersections) we need all open rectangles to be open. But this doesn't make a topology yet, as we close it under unions too. And we can check open rectangles can form a base for a topology so their union form a unique topology (which must be the minimal one making all projections continuous, because that already forces these sets to be in the topology), and so is the required product topology.