Proving that the product of the elements of the bases are elements of a basis for the box topologies I want to prove the following theorem:

If the topology on each space $X_\alpha$ is generated by a basis $\mathcal{B}_{\alpha}$. The set $\mathcal{B}={\underset{\alpha\in J}{\prod}B_{\alpha}:\forall\alpha\in J,\,B_{\alpha}\in\mathcal{B_{\alpha}}}
  $ is therefore a basis for the box topology on $\underset{\alpha\in J}{\prod}X_{\alpha}
 $.

My attempt: Let $U$ be an open set in the box topology of $\underset{\alpha\in J}{\prod}X_{\alpha}
 $. I want to prove that it can be written as the union of elements of $\mathcal{B}
 $. We can write $U$ as the union of elements of the standard basis for the box topology: $U=\underset{i\in I}{\bigcup}\underset{\alpha\in J}{\prod}U_{\alpha}^{i}
 $ (with each $U_{\alpha}^{i}
 $ an open set of $X_\alpha$). Now, each $U_{\alpha}^{i}
 $ can be written as the union of elements of $\mathcal{B}_{\alpha}
 $: $U_{\alpha}^{i}=\underset{\lambda\in\Lambda}{\bigcup}B_{\alpha,\lambda}^{i}
 $. Thus, we get $U=\underset{i\in I}{\bigcup}\underset{\alpha\in J}{\prod}\underset{\lambda\in\Lambda}{\bigcup}B_{\alpha,\lambda}^{i}=\underset{i\in I}{\bigcup}\underset{\lambda\in\Lambda}{\bigcup}(\underset{\alpha\in J}{\prod}B_{\alpha,\lambda}^{i})
 $. The factor in parentheses is an element of $\mathcal{B}
 $.
I don't know how to get $U$ as an union of such elements of $\mathcal{B}$  in order to conclude the proof.
 A: A common method for proving that a set $U$ is a union of elements of a collection $\mathcal B$ is to prove that for each point $x \in U$ there exists $B \in \mathcal B$ such that $x \in B \subset U$.
So suppose that $x \in U$. In more detail, $x=(x_\alpha)_{\alpha \in J} \in U = \bigcup_{i \in I} \prod_{\alpha \in J} U^i_\alpha$. 
Then there exists $i \in I$ such that $x \in \prod_{\alpha \in J} U^i_\alpha$. 
It follows that for each $\alpha \in J$ we have $x_\alpha \in U^i_\alpha$. 
Because $\mathcal B_\alpha$ is a basis for $X_\alpha$, it follows that there exists $B^i_\alpha \in \mathcal{B}_\alpha$ such that $x_\alpha \in B^i_\alpha \subset U^i_\alpha$. 
Therefore $x \in B^i \equiv \prod_{\alpha \in J} B^i_\alpha \subset \prod_{\alpha \in J} U^i_\alpha \subset U$, and $B^i \in \mathcal{B}$.
A: As concerns your attempt, the family $\Lambda$ such that
$U_\alpha^i=\bigcup_{\lambda\in\Lambda}B^i_{\alpha,\lambda}$
depends on $\alpha$, so it is appropriate to denote it $\Lambda_\alpha$.
Then you can see that it has no sense to write
$\bigcup_{\lambda\in\Lambda_{\alpha}}\big(\prod_{\alpha\in J}B^i_{\alpha,\lambda}\big)$, as you did.
So, your last equality is not valid.
The term $\prod_{\alpha\in J}\bigcup_{\lambda\in\Lambda_\alpha} B^i_{\alpha,\lambda}$
denotes the family of all functions $f$ with domain $J$ such that
$f(\alpha)$ is an element of $\bigcup_{\lambda\in\Lambda_\alpha}B^i_{\alpha,\lambda}$, for each $\alpha\in J$.
For any such $f$ we can find a function $g$ mapping each $\alpha\in J$ to $g(\alpha)\in\Lambda_\alpha$ such that
$f(\alpha)\in B^i_{\alpha,g(\alpha)}$,
thus $g\in\prod_{\alpha\in J}\Lambda_\alpha$ and $f\in\prod_{\alpha\in J} B^i_{\alpha,g(\alpha)}$.
Also, for any $g\in\prod_{\alpha\in J}\Lambda_\alpha$, each $f\in\prod_{\alpha\in J} B^i_{\alpha,g(\alpha)}$
is an element of $\prod_{\alpha\in J}\bigcup_{\lambda\in\Lambda_\alpha} B^i_{\alpha,\lambda}$.
So we can write
$$\prod_{\alpha\in J}\bigcup_{\lambda\in\Lambda_\alpha} B^i_{\alpha,\lambda}=
\bigcup_{g\in\prod_{\alpha\in J}\Lambda_\alpha}\quad\prod_{\alpha\in J} B^i_{\alpha,g(\alpha)},$$
which is a union of elements of $\mathcal{B}$.
