Let $(X,\mathcal O_X), (Y, \mathcal O_Y)$ be two topological spaces, and let $\mathcal B=\{ U\times V \mid U\in \mathcal O_X, V\in \mathcal O_Y \}$.
Product topology $\mathcal O$ is a family of sets that is composed of $\mathcal B$,i.e., $\mathcal O=\langle \mathcal B \rangle$.
I want to prove $\mathcal O$ is the minimum topology such that both projections $f_1 : X\times Y \to X, f_2 : X\times Y \to Y$ are continuous.
I proved that $\mathcal O$ is a topology and $f_1 : (X\times Y, \mathcal O)\to (X, \mathcal O_X)$ and $f_2 : (X\times Y, \mathcal O)\to (Y, \mathcal O_Y)$ are continuous.
I have to prove the minimality of $\mathcal O$.
What I have to show is that if $\mathcal O'$ is a topology of $X\times Y$ s.t. $f_1 : (X\times Y, \mathcal O')\to (X, \mathcal O_X)$ and $f_2 : (X\times Y, \mathcal O')\to (Y, \mathcal O_Y)$ are continuous, then $\mathcal O \subset \mathcal O'$.
Let $O\in \mathcal O.$
Then, since $\mathcal O=\langle \mathcal B \rangle$, we can write $O=\cup_{\lambda \in \Lambda}O_\lambda$ where $\Lambda$ is some index set, and each $O_\lambda$ belongs to $\mathcal B.$
Since each $O_\lambda$ belongs to $\mathcal B$, there are $U_\lambda \in \mathcal O_X$, $V_\lambda\in \mathcal O_Y$ s.t. $O_\lambda=U_\lambda \times V_\lambda.$
Then, $O=\cup_{\lambda\in \Lambda} O_\lambda=\cup_{\lambda\in \Lambda} (U_\lambda \times V_\lambda)$.
I want to show this belongs to $\mathcal O'$, but I don't know how.
I know I have to use the continuity of $f_1, f_2.$
I have $f_1^{-1}(\cup U_\lambda)=\cup f_1^{-1}(U_\lambda)=\cup (U_\lambda \times Y)$, similarly, $f_2^{-1}(\cup V_\lambda)=\cup (X\times V_\lambda).$ And by the continuity, $f_1^{-1}(\cup U_\lambda), f_2^{-1}(\cup V_\lambda)\in \mathcal O'$.
$\mathcal O' \ni f_1^{-1}(\cup U_\lambda)\cap f_2^{-1}(\cup V_\lambda) =\Big(\cup (U_\lambda \times Y) \Big)\cap \Big(\cup (X\times V_\lambda)\Big)$.
Therefore, if I can show $O=\cup_{\lambda\in \Lambda} (U_\lambda \times V_\lambda)=\Big(\cup_{\lambda\in \Lambda} (U_\lambda \times Y) \Big)\cap \Big(\cup_{\lambda\in \Lambda} (X\times V_\lambda)\Big)$, I can say $O\in \mathcal O'.$
However, I don't think $\cup_{\lambda\in \Lambda} (U_\lambda \times V_\lambda)=\Big(\cup_{\lambda\in \Lambda} (U_\lambda \times Y) \Big)\cap \Big(\cup_{\lambda\in \Lambda} (X\times V_\lambda)\Big)$, holds.
$\subset$ is obvious, but I think $\supset$ doesn't hold.
Is there a point where I make a mistake ? How can I show $\mathcal O\subset \mathcal O'$ ?
(1). If $B$ is a collection of subsets of a set $Z$ then $B$ is a base (basis) for a topology on $Z$ iff (i) $\cup B=Z$ (i.e. each $p\in Z$ belongs to at least one member of $B$) and (ii) whenever $b_1,b_2\in B$ and $p\in b_1\cap b_2,$ there exists $b_3\in B$ with $p\in b_3\subseteq b_1\cap b_2.$
If $b_1\cap b_2\in B$ whenever $b_1,b_2\in B$ then (ii) (above) is satisfied by letting $b_3=b_1\cap b_2.$
From this it is easy to see that $\mathcal B$ is a base for a topology $T_{\mathcal B}$ on $X\times Y.$
(2). If $B$ is a base for a topology $T$ on a set $Z,$ and if $T'$ is a topology on $Z$ with $T'\supseteq B$ then $T'\supseteq T. $
So it suffices to show that if $T'$ is a topology on $X\times Y$ such that $f_1,f_2$ are continuous then $T'\supseteq \mathcal B,$ as follows:
For any $U\in \mathcal O_X$ and any $V\in \mathcal O_Y$ we have $U\times Y=f_1^{-1}U\in T'$ and $X\times V=f_2^{-1}V\in T'.$ Therefore $U\times V=(f_1^{-1}U)\cap (f_2^{-1}V)\in T'.$