Ok, so we have to prove Theorem 19.2, which asks us to consider a collection of topological spaces indexed by $\alpha$, called $X_{\alpha}$. Now, each topological space
has a basis given by $\beta_{\alpha}$. We need to show that the collection of all sets of the form:
$$\displaystyle\prod_{\alpha \ \in \ J} \ B_{\alpha}$$
Forms a basis for the box topology on $\prod_{a \ \in \ J} \ X_{\alpha}$, where $B_{\alpha} \ \in \ \beta_{\alpha}$. In addition, the collection of the same form
for $B_{\alpha} \ \in \ \beta_{\alpha}$ for a finite number of $\alpha$, and $B_{\alpha} \ = \ X_{\alpha}$ will serve as a basis for the product topology on $\prod_{\alpha \ \in \
J} \ X_{\alpha}$. Let's start with the box topology. Let us pick some point $x_{\alpha}$, such that $x_{\alpha} \ \in \ B_{\alpha} \ \subset \ U_{\alpha}$. This means that some
element $\textbf{x} \ = \ (x_{\alpha})$, we will have:
$$(x_{\alpha}) \ \in \ \displaystyle\prod_{\alpha \ \in \ J} \ B_{\alpha} \ \subset \ \displaystyle\prod_{\alpha \ \in \ J} \ U_{\alpha}$$
Since each "factor" of the tuple $(x_{\alpha})$ is contained within its respective basis, which is contained within its respective open set. A similar argument applies for the
product topology. For each of our finite basis elements, we again have:
$$(x_{\alpha}) \ \in \ \displaystyle\prod_{\alpha \ \in \ J} \ B_{\alpha} \ \subset \ \displaystyle\prod_{\alpha \ \in \ J} \ U_{\alpha}$$
But this time, for non-finite indices, we have $B_{\alpha} \ = \ X_{\alpha}$ and $U_{\alpha} \ = \ X_{\alpha}$. It makes sense that $\beta_{\alpha}$ is a basis,
as for indices, except for the finite ones, we have $x_{\gamma} \ \in \ X_{\gamma} \ \subset \ X_{\gamma}$, which is the definition of a basis element $B_{\gamma} \ = \ X_{\gamma}$.