Why does the product topology allow proper subsets for only finitely many elements? Consider Theorem 19.1 from Munkres' topology:

The box topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$.

I am having a hard time seeing how the last sentence follows from the subbasis-definition of the product topology. I don't find Munkres' preceding explanation to be helpful.

There are two ways to view the generation of a topology from a subbasis $\mathcal S$.


*

*The topology generated by $\mathcal S$ is the smallest (coarsest) topology in which all the sets in $\mathcal S$ are open.

*We first transform $\mathcal S$ into a basis $\mathcal B$ consisting of all (nonempty) finite intersections of sets in $\mathcal S$, and then take the topology generated by this basis.
These two ways can be shown to be equivalent, though the second method is somewhat more illustrative in the current situation, since we outright say that the basis open sets are finite intersections of subbasic open sets. This means that $U$ is open in the generated topology if and only if for each $x \in U$ there are $W_1 , \ldots , W_n$ (for some $n \geq 1$) in $\mathcal S$ such that $x \in W_1 \cap \cdots \cap W_n \subseteq U$.
Since the subbasis for the product topology is the family of all sets $\pi_\beta^{-1} ( U )$ where $\beta$ an index, and $U$ is open in $X_\beta$, then the basic open sets are of the form $$\pi_{\beta_1}^{-1} ( U_{\beta_1} ) \cap \cdots \cap \pi_{\beta_n}^{-1} ( U_{\beta_n} ),$$ where $\beta_1 , \ldots , \beta_n$ are indexes, and $U_{\beta_i}$ is open in $X_{\beta_i}$ for all $i$.  We may then note that if $\beta_i = \beta_j = \beta$, then $\pi_{\beta_i}^{-1} ( U_{\beta_i} ) \cap \pi_{\beta_j}^{-1} ( U_{\beta_j} ) = \pi_{\beta}^{-1} ( U_{\beta_i} \cap U_{\beta_j} )$, and $U_{\beta_i} \cap U_{\beta_j}$ is open in $X_\beta$, and so each index need only appear (at most) once.
