The collection $$\mathcal{S} = \{U\times Y |U \text{ open in X}\} \cup \{X\times V|V \text{ open in Y}\}$$ is a subbasis for the product topology on $X \times Y$.
I’m not using fancy inverse pie notation.
My attempt: (1) first I will show $\mathcal{S}$ is a subbasis. First condition is to show $\mathcal{S}\subseteq \wp({X\times Y})$. Let $M\in \mathcal{S}$. $M\in \{U\times Y |U \text{ open in X}\}$ or $M\in \{X\times V|V \text{ open in Y}\}$. If $M\in \{U\times Y |U \text{ open in X}\}$, then $M=N\times Y$, $N\in \mathfrak{I}_{X}\subseteq \wp({X})$. $N\in \wp({X})$. So, $N\subseteq X$. Thus $M=N\times Y \subseteq X\times Y$, $M\in \wp({X\times Y})$. Similar proof applies if $M\in \{X\times V|V \text{ open in Y}\}$. Thus $\mathcal{S}\subseteq \wp({X\times Y})$. Now we’ll show the second condition $\cup \mathcal{S}=X\times Y$. If $M\in \cup \mathcal{S}$, then $\exists S\in \mathcal{S}$ such that $M\in S\subseteq X\times Y$. So $M\in X\times Y$. Thus $\cup \mathcal{S} \subseteq X\times Y$. Conversely, let $M\in X\times Y$. By definition of basis, $\exists R\times S \in \mathscr{B}_{p}$(is a basis defined in definition of section 15) such that $M\in R\times S\subseteq R\times Y$. The last inclusion follows from, $S\in \mathscr{I}_{Y} \subseteq \wp({Y})$. So $S\subseteq Y$. Note $R\times Y \in \mathcal{S}$. Thus $M\in R\times Y \subseteq \cup \mathcal{S}$. Hence $\cup \mathcal{S}=X\times Y$. Is this proof correct?
(2) I want to show $\mathcal{T}\subseteq \mathcal{T}^\prime$, where $\mathcal{T}$ denote the product topology on $X\times Y$ and $\mathcal{T}^\prime$ denotes topology generated by $\mathcal{S}$. My solution is different from Munkres. Let $U\times V \in \mathcal{T}$. By the definition of topology generated by basis, $\forall (u,v)\in U\times V, \exists W_{(u,v)} \in \mathscr{B}_{p}\subseteq$ $\mathcal{B}$(basis of $\mathcal{S}$. Subset holds because $(A\times Y)\cap (X\times B)=A\times B$) such that $(u,v)\in W_{(u,v)} \subseteq U\times V$. so $U\times V = \cup_{(u,v)\in U\times V} W_{(u,v)}$. $U\times V \in \mathcal{T}^\prime$. Thus $\mathcal{T}\subseteq \mathcal{T}^\prime$. Is this proof correct?
To prove this inclusion. I think Munkres used lemma 13.3. Constructed the basis of $\mathcal{T}^\prime$ by finite intersection of elements of $\mathcal{S}$.
Edit: Conversation between Henno Brandsma and I has been moved into chat. According to me, those conversation are more interesting and useful than the post itself. Most people(who are searching this theorem on google) are not going to go in the chat room to read our comments. So I’m adding those comments.
User264745: Thank you so much for the answer. To summarize this entire(your and mine) post: (1) There are two ways(though your way is much shorter than mine) to show $\mathcal{S}$ is a subbasis. (2) There are three way to show topology generated by subbasis $\mathcal{S}$ is equal to product topology on $X\times Y$. 1) first is Munkres or Henno approach, basis for product topology = basis generated from subbasis, though you could have taken general finite intersection of elements of $\mathcal{S}$ and show it is one of the point in $\mathcal{B}$.2) The way I showed above, arbitrary union of finite intersection of elements of $\mathcal{S}$. 3) using lemma 13.3, maybe before using that lemma we have to show set $\mathcal{B}$ is indeed a basis for product topology, to do that one have show $\mathcal{B}$ satisfy the hypothesis of lemma 13.2. *one of the point in standard basis for $\mathcal{T}_p$.
Henno: I don't think 13.3 is the right tool here. It's overkill, and only shows one inclusion of topologies anyway. Realise what a subbase is and you're led to my proof above. Taking arbitrary finite intersections adds nothing to my proof. We already know the standard base for $\mathcal{T}_p$ is included in the base generated by $\mathcal{S}$ and we already know all members of $\mathcal S$ are standard open base sets. The standard base for $\mathcal T_p$ is "special" already in that it's already closed under finite intersections, which need not be the case for all bases.
User264745: I’m sorry, I didn’t understand your last few lines of reasoning. I talked about general intersection because of the following reasons. Our goal is to show $\mathscr{B}_{p}$(basis of product topology) = $\mathcal{B}$(basis generated by $\mathcal{S}$). Proof: $\mathscr{B}_{p} \subseteq \mathcal{B}$ is pretty clear. Conversely, if $U\times V \in \mathcal{B}$ ,then $U\times V=(U_1 \times V_1) \cap \dotsb \cap (U_n \times V_n)= (U_1 \cap \dotsb \cap U_n) \times (V_1 \cap \dotsb \cap V_n)$. I’m not writing $U_1$ or $V_1$ as $X$ or $Y$, for seek of generality.$U_1 \cap \dotsb \cap U_n \in \mathcal{T}_{X}$ and $V_1 \cap \dotsb \cap V_n \in \mathcal{T}_{Y}$. Thus $U\times V \in \mathscr{B}_{p}$. Hence $\mathscr{B}_{p} \supseteq \mathcal{B}$.
User264745: Using lemma 13.3 is overkill, for sure. We have shown two basis are equal so we don’t have to show $\mathcal{T} \subseteq \mathcal{T}_p$.
Henno: $\mathcal S\subseteq \mathcal B_p$ and the latter base is closed under finite intersection so $\mathcal B \subseteq \mathcal B_p$ follows. The other inclusion is trivial QED.
User264745: But the basis $\mathcal{B}_p$ is not a Topology on $X\times Y$.
Henno: But it is closed under finite intersections as $\bigcap_{i=1}^n (U_i \times V_i) = \bigcap_{i=1}^n U_i \times \bigcap_{i=1}^n V_i$ by simple set theory.
User264745: Ohh.. So we’re using this fact implicitly.
Henno: every thing from set theory is always supposed known and usable. Ditto for logic and elementary number facts.
User264745: Yeah I mean, if someone didn’t notice/know this property, would take bigger route(like I did). In the process of that route, one have to prove, the so called closed under finite intersection property.
User264745: Hey! I have found another method(efficient/ shortest) to solve this problem. The following approach is inspired by previous theorem(15.1). Let $\mathcal{B}$ is the basis generated from $\mathcal{S}$. If we can show $\mathcal{B}$ is the basis for product topology then $\mathcal{T}_{\mathcal{B}}= \mathcal{T}_{p}= \mathcal{T}_{\mathcal{S}}$(last topology is generated by subbasis). Proof: first condition $\mathcal{B}\subseteq \mathcal{T}_{p}$. Second condition, let $U\times V\in \mathcal{T}_p$ and $(u,v)\in U\times V,\exists (M,N)\in \mathcal{B}$ s.t $(u,v)\in (M,N)\subseteq (U,V).$
I hope, I didn’t make any mistake in excitement. I used $(A,B)$, because of word constraint.
Nahhhh..... We somehow use $\mathcal{B}_p = \mathcal{B}$ fact.
You're making this way too hard. Let $\mathcal{T}$ be the topology generated by the subbase $\mathcal{S}$, and let $\mathcal{T}_p$ be the product topology on $X \times Y$.
All members of $\mathcal{S}$ are subsets of $X \times Y$ and in particular we can see that $X \times Y \in \mathcal{S}$ (we can choose $U=X$ in the first type, and also $V=Y$ in the second type). So that $\mathcal{S}$ is a subbase (in the Munkres sense) is totally obvious.
All members of $\mathcal{S}$, be they of the form $U \times Y$ or $X \times V$ are in $\mathcal{T}_p$ by definition. So trivially $\mathcal{T} \subseteq \mathcal{T}_p$. And if we look at the base $\mathcal{B}$ generated by $\mathcal{S}$, we get all sets $(U \times Y) \cap (X \times V) = U \times V$ in $\mathcal{B}$ so that the base generated by $\mathcal{S}$ is precisely the standard base for $\mathcal{T}_p$ so we have equality of these topologies. That's really all you have to realise.
Two topologies with the same base (basis if you prefer) are equal...