Theorem: Suppose the topology on each space $X_{\alpha}$ is given by a basis $\mathcal{B}_{\alpha}$. The collection of all sets of the form $$\prod_{\alpha \in J} B_{\alpha}$$ where $B_{\alpha} \in \mathcal{B}_{\alpha}$ for each ${\alpha}$, forms a basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$
My Attempted Proof
The topological space we are dealing with here is $\left(\prod_{\alpha \in J} X_{\alpha}, \mathcal{T}_B\right)$, where $\mathcal{T}_B$ is the box topology on $\prod_{\alpha \in J} X_{\alpha}$. Let $\mathcal{B}_{\text{box}}$ denote the standard basis for the box topology.
Put $$\mathcal{F} = \left\{ \ \prod_{\alpha \in H} B_{\alpha} \ \middle| \ B_{\alpha} \in \mathcal{B}_{\alpha} \ \text{for each $\alpha$} \ \right\}.$$ Then take $U \in \mathcal{T}_B$, and take $x \in U$. Since $U = \bigcup_{ B \in \mathcal{K}}B$ where $B = \prod_{\alpha \in J}U_{\alpha}$ and $U_{\alpha}$ is open in $X_{\alpha}$ and $\mathcal{K} \subset \mathcal{B}_{\text{box}}$ we then have $x \in B$ for some $B \in \mathcal{K}$. In other words $x \in \prod_{\alpha \in J}U_{\alpha}$ for some indexing set $J$.
Now since each $U_{\alpha}$ is open in $X_{\alpha}$, we have $U_{\alpha} = \bigcup_{\gamma \in I}B_{\gamma}$ where $B_{\gamma}$ are basis elements of $X_{\alpha}$. Therefore $$x \in \prod_{\alpha \in J}\left(\bigcup_{\gamma \in I}B_{\gamma}\right)_{\alpha}$$
By elementary set theory we have $$\left(\bigcup_{i \in I} V_i\right) \times \left(\bigcup_{j \in J} W_j\right) = \bigcup_{\langle i, j \rangle I \times J}\left(V_i \times W_j\right)$$
So that $$\prod_{\alpha \in J}\left(\bigcup_{\gamma}B_{\gamma}\right)_{\alpha} \ = \ \bigcup_{\langle i_1, i_2, ... \rangle \in I_1 \times I_2 \times ...} \underbrace{\left(B_{i_1} \times B_{i_2} \times ...\right)}_{|J| \ \text{times}}$$
Choose $H = I_1 \times I_2 \times I_3 \times ...$
Then $\exists \prod_{\alpha \in H}B_{\alpha} \in \mathcal{F}$ such that $x \in \prod_{\alpha \in H}B_{\alpha}$. It also follows that $$\prod_{\alpha \in J}\left(\bigcup_{\gamma}B_{\gamma}\right)_{\alpha} \subset U$$ Since $\prod_{\alpha \in J}\left(\bigcup_{\gamma}B_{\gamma}\right)_{\alpha} = \prod_{\alpha \in J}U_{\alpha} = B \subset U$. Thus it follows that $\mathcal{F}$ is a basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$. $\square$
Is my above proof correct and rigorous? Are there areas of it that I can improve on? Can this be proved in an easier way? I apologize in advance if the proof is long or messy (especially with the indices)
If my proof is incorrect however, or if it is not rigorous please let me know!
Suppose $x =(x_i)_i \in O$, where $O$ is box open. This means that there are open sets $O_i \subseteq X_i$, such that $x \in \prod_i O_i \subseteq O$, by the definition of the box topology (topology generated by open "boxes", so the latter form a base). As each $\mathcal{B}_i$ is a base in $X_i$, we pick $B_i \in \mathcal{B}_i$ such that $x_i \in B_i \subseteq O_i$, but then $x \in \prod_i B_i (\in \mathcal{F}) \subseteq \prod_i O_i \subseteq O$. So sets from $\mathcal{F}$ form a base.
There is no need for set-theory manipulation here, just standard definitions.