I realize these two topologies have been discussed before, but I can't understand why the product topology's components are the spaces $X_\alpha$ except for finitely many components. It is probably a silly mistake. In Munkres' text this is a theorem, not part of the definition. He gives the following definitions:
Let $\{X_\alpha\}_{\alpha \in J}$ be an indexed family of topological spaces. Let us take as a basis for a topology on the product space $$\prod_{\alpha\in J} X_\alpha$$ the collection of all sets of the form $$\prod_{\alpha\in J} U_\alpha$$ where $U_\alpha$ is open in $X_\alpha$ for each $\alpha\in J$. The topology generated by this basis is called the box topology.
Then he defines the projection $\pi_\beta$ as usual for $\beta\in J$. Then he defines the product topology:
Let $\mathcal{S}_\beta$ denote the collection $$\mathcal{S}_\beta = \{ \pi^{-1}(U_\beta) | U_\beta \> \text{open in}\ \> X_\beta \},$$ and let $\mathcal{S}$ denote the union of these collections $$\mathcal{S} = \bigcup_{\beta\in J}\mathcal{S}_\beta$$ The topology generated by the subbasis $\mathcal{S}$ is called the product topology.
I see two possibilities for the interpretation of the definition of the set $\mathcal{S}_\beta$, and in particular the status of $U_\beta$. In one interpretation, for each $\beta$, $U_\beta$ is a fixed open set in $X_\beta$, so that $\mathcal{S}_\beta$ consists of all products of the form $$A_{\alpha_1} \times A_{\alpha_2} \times \cdots \times U_\beta \times \cdots$$
where for each $\alpha_i$ the $A_{\alpha_i}$ are any open set in $X_{\alpha_i}$, and $U_\beta$ is in the '$\beta$th' component. This is just the set of products which under the projection map to $U_\beta$, i.e. it is the preimage.
The other interpretation is that for each $\beta$, $U_\beta$ is 'universally quantified' over all open sets in $X_\beta$, making each product in $\mathcal{S}_\beta$ of the form $$A_{\alpha_1} \times A_{\alpha_2} \times \cdots \times U_{j,\beta} \times \cdots$$ where $U_{j,\beta}$ is any open set in $X_\beta$, indexing the open sets of $X_\beta$ by $j$ for clarity, so that the open sets of $X_\beta$ are exactly $\{ U_{j,\beta}\}$ (and of course $U_{j,\beta}$ in the $\beta$th component as before). However my understanding is that this would mean that for each $\beta$, $\mathcal{S}_\beta$ is equal to the box product, so I assume that is not the correct interpretation.
However, even if the first interpretation is correct, this means that $\mathcal{S}$ is the entire box product, because its elements are of the form
$$\mathcal{A} = U_{\alpha_1} \times U_{\alpha_2} \times \cdots, $$where each $U_{\alpha_i}$ is universally quantified over the open sets in $X_{\alpha_i}$. And in particular, for such elements $\mathcal{A}\cap\mathcal{B}$, we have $$(U_{\mathcal{A},\alpha_1}\cap U_{\mathcal{B},\alpha_1}) \times (U_{\mathcal{A},\alpha_1}\cap U_{\mathcal{B},\alpha_2}) \times \cdots,$$ where since each component is an open set in $X_{\alpha_i}$, this intersection is again contained in $\mathcal{S}$ by my assumption.
I understand that something here, or quite a bit, is incorrect, but I don't see what, and in general I don't see how to prove that the product topology has for open sets the products with only finitely many components $\beta$ distinct from their space $X_\beta$. I suspect that the finiteness of the intersections performed on a subbasis somehow manifest into this finitely-many-proper-subset-components property, but I can't get past the definitions discussed here.
Best Answer
There is a slight misunderstanding here about the product (and box) topologies. It is not true that every open set (in either topology) is a product of open sets. Indeed, consider the open unit ball $\{(x,y)\in\mathbb{R}^2 \mid x^2+y^2< 1\}$ in $\mathbb{R}^2$. This set is open in both topologies (because they coincide for finite products), but it is not of the form $U\times V$ with $U,V\subseteq\mathbb{R}$ open. The key fact is that these products form a basis for the topology.
Recall that if $B$ is a base which generates the topology $(X,\tau)$, then $$ \tau = \{U\subseteq X \mid U\ \text{is a union of elements of $B$}\}. $$
This is equivalent to saying that $U\subseteq X$ is open iff for every $x\in U$ there exists $V\in B$ such that $x\in V \subseteq U$. (Try justifying why these two are equivalent.)
On the other hand, we can consider a subbase. If $S$ is a subbase which generates the topology $(X,\tau)$, then $$ \tau = \{U\subseteq X \mid U\ \text{is a union of finite intersections of elements of $S$}\}. $$ Note also that if $S$ is a subbase, then we can obtain a base via $$ B = \{V\subseteq X \mid V\ \text{is a finite intersection of elements of $S$}\}. $$ This gets to the heart of a matter. We want to prove:
We can do this with the process described above to turn a subbase into a base. We know that the set $$ S = \{\pi_i^{-1}(U_i) \mid i\in I,\, U_i\subseteq X_i\ \text{is open}\} $$ is a subbase for the product topology. This means that sets of the form $$ \pi_{i_1}^{-1}(U_{i_1})\cap \cdots \cap \pi_{i_n}^{-1}(U_{i_n}), $$ where $i_1,\ldots,i_n\in I$ and $U_{i_k}\subseteq X_{i_k}$ is open for each $k=1,\ldots,n$, forms a base for the topology. Since the preimage respects unions and intersections, we can assume that $i_1,\ldots,i_n$ are distinct. Then we have (why?) $$ \pi_{i_1}^{-1}(U_{i_1})\cap \cdots \cap \pi_{i_n}^{-1}(U_{i_n}) = \prod_{i\in I}U_i, $$ where $$ U_i = \begin{cases} U_{i_k} & \text{if $i=i_k$ for some $k=1,\ldots,n$},\\ X_i & \text{otherwise}. \end{cases} $$