[proofplan]
We compare two topologies on the same set $M \times N$: the [product topology](/page/Product%20Topology) coming from the manifold topologies on $M$ and $N$, and the topology induced by the standard atlas of product charts. The first inclusion follows because every chart-[open set](/page/Open%20Set) is locally pulled back from an open set in Euclidean space, and Euclidean product coordinates have product-open neighbourhood bases. The reverse inclusion follows by checking a product-basic open set $A \times B$ in every product chart.
[/proofplan]
[step:Define the two topologies to be compared]
Let $m,n \in \mathbb{Z}_{\ge 0}$ be the dimensions of $M$ and $N$, respectively. Let $\tau_{\mathrm{prod}}$ denote the product topology on $M \times N$ generated by sets $A \times B$, where $A \in \tau_M$ and $B \in \tau_N$.
Let $\mathcal{A}$ denote the standard product atlas on $M \times N$. Its charts are the maps $\Phi_{\varphi,\psi}: U \times V \to \varphi(U) \times \psi(V) \subseteq \mathbb{R}^{m+n}$ defined by $\Phi_{\varphi,\psi}(p,q) = (\varphi(p),\psi(q))$, where $(U,\varphi)$ is a smooth chart on $M$ and $(V,\psi)$ is a smooth chart on $N$. Here $\mathbb{R}^m \times \mathbb{R}^n$ is identified with $\mathbb{R}^{m+n}$ by the map $I: \mathbb{R}^m \times \mathbb{R}^n \to \mathbb{R}^{m+n}$ given by $I((x_1,\dots,x_m),(y_1,\dots,y_n)) = (x_1,\dots,x_m,y_1,\dots,y_n)$.
Let $\tau_{\mathcal{A}}$ denote the topology induced by this atlas: a set $W \subseteq M \times N$ is in $\tau_{\mathcal{A}}$ exactly when, for every product chart $\Phi_{\varphi,\psi}: U \times V \to \varphi(U) \times \psi(V)$, the image $\Phi_{\varphi,\psi}(W \cap (U \times V))$ is open in $\varphi(U) \times \psi(V)$ with its Euclidean [subspace topology](/page/Subspace%20Topology).
[/step]
[step:Show every product-chart domain is open in the product topology]
Let $(U,\varphi)$ be a chart on $M$ and $(V,\psi)$ a chart on $N$. Since $U \in \tau_M$ and $V \in \tau_N$ by the definition of a manifold chart, the set $U \times V$ belongs to $\tau_{\mathrm{prod}}$ by the definition of the product topology. Hence every product-chart domain of $\mathcal{A}$ is open in $\tau_{\mathrm{prod}}$.
[/step]
[step:Prove that every atlas-open set is product-open]
Let $W \in \tau_{\mathcal{A}}$, and let $(p,q) \in W$. Choose a chart $(U,\varphi)$ on $M$ with $p \in U$ and a chart $(V,\psi)$ on $N$ with $q \in V$. Define the product chart $\Phi_{\varphi,\psi}: U \times V \to \varphi(U) \times \psi(V) \subseteq \mathbb{R}^{m+n}$ by $\Phi_{\varphi,\psi}(r,s) = (\varphi(r),\psi(s))$. Because $W \in \tau_{\mathcal{A}}$, the set $\Phi_{\varphi,\psi}(W \cap (U \times V))$ is open in $\varphi(U) \times \psi(V)$. Since $\Phi_{\varphi,\psi}(p,q)$ lies in this open set and the Euclidean topology on $\mathbb{R}^{m+n}$ agrees under $I$ with the product topology on $\mathbb{R}^m \times \mathbb{R}^n$, there exist open sets $O \subseteq \varphi(U)$ and $P \subseteq \psi(V)$ such that
\begin{align*}
\varphi(p) \in O, \qquad \psi(q) \in P, \qquad O \times P \subseteq \Phi_{\varphi,\psi}(W \cap (U \times V)).
\end{align*}
The chart maps $\varphi: U \to \varphi(U)$ and $\psi: V \to \psi(V)$ are homeomorphisms, so $\varphi^{-1}(O)$ is open in $U$ and $\psi^{-1}(P)$ is open in $V$. Since $U \in \tau_M$ and $V \in \tau_N$, the sets $\varphi^{-1}(O)$ and $\psi^{-1}(P)$ are open in $M$ and $N$, respectively. Therefore
\begin{align*}
\varphi^{-1}(O) \times \psi^{-1}(P) \in \tau_{\mathrm{prod}}.
\end{align*}
This product neighbourhood contains $(p,q)$ and is contained in $W$. Since every point of $W$ has a $\tau_{\mathrm{prod}}$-open neighbourhood contained in $W$, we have $W \in \tau_{\mathrm{prod}}$. Thus $\tau_{\mathcal{A}} \subseteq \tau_{\mathrm{prod}}$.
[guided]
We want to prove that an arbitrary atlas-open set $W$ is open in the product topology. The product topology is local: it is enough to show that each point of $W$ has a product-open neighbourhood contained in $W$.
Fix $(p,q) \in W$. Choose a chart $(U,\varphi)$ on $M$ with $p \in U$ and a chart $(V,\psi)$ on $N$ with $q \in V$. The corresponding product chart is the map $\Phi_{\varphi,\psi}: U \times V \to \varphi(U) \times \psi(V) \subseteq \mathbb{R}^{m+n}$ given by $\Phi_{\varphi,\psi}(r,s) = (\varphi(r),\psi(s))$. Because $W$ is open for the topology induced by the atlas, the coordinate image $\Phi_{\varphi,\psi}(W \cap (U \times V))$ is open in the Euclidean subspace $\varphi(U) \times \psi(V)$.
The point $\Phi_{\varphi,\psi}(p,q) = (\varphi(p),\psi(q))$ lies in that open coordinate image. The Euclidean topology on $\mathbb{R}^{m+n}$ is the same as the product topology on $\mathbb{R}^m \times \mathbb{R}^n$ under the coordinate identification $I((x_1,\dots,x_m),(y_1,\dots,y_n)) = (x_1,\dots,x_m,y_1,\dots,y_n)$. Therefore there are open sets $O \subseteq \varphi(U)$ and $P \subseteq \psi(V)$ such that
\begin{align*}
\varphi(p) \in O, \qquad \psi(q) \in P, \qquad O \times P \subseteq \Phi_{\varphi,\psi}(W \cap (U \times V)).
\end{align*}
Now pull this Euclidean product neighbourhood back to $M \times N$. Since $\varphi: U \to \varphi(U)$ and $\psi: V \to \psi(V)$ are homeomorphisms, $\varphi^{-1}(O)$ is open in $U$ and $\psi^{-1}(P)$ is open in $V$. Because $U$ and $V$ are themselves open in the manifold topologies, these inverse images are open in $M$ and $N$. Hence
\begin{align*}
\varphi^{-1}(O) \times \psi^{-1}(P)
\end{align*}
is open in the product topology on $M \times N$.
Finally, the containment $O \times P \subseteq \Phi_{\varphi,\psi}(W \cap (U \times V))$ implies
\begin{align*}
\varphi^{-1}(O) \times \psi^{-1}(P) \subseteq W.
\end{align*}
Thus every point of $W$ has a product-open neighbourhood contained in $W$, so $W$ is product-open. This proves $\tau_{\mathcal{A}} \subseteq \tau_{\mathrm{prod}}$.
[/guided]
[/step]
[step:Prove that every product-basic open set is atlas-open]
Let $A \in \tau_M$ and $B \in \tau_N$. We prove that $A \times B \in \tau_{\mathcal{A}}$. Let
\begin{align*}
\Phi_{\varphi,\psi}: U \times V \to \varphi(U) \times \psi(V) \subseteq \mathbb{R}^{m+n}
\end{align*}
be an arbitrary product chart. Then
\begin{align*}
(A \times B) \cap (U \times V) = (A \cap U) \times (B \cap V).
\end{align*}
Applying the product chart gives
\begin{align*}
\Phi_{\varphi,\psi}((A \times B) \cap (U \times V)) = \varphi(A \cap U) \times \psi(B \cap V).
\end{align*}
Since $A \cap U$ is open in $U$ and $\varphi: U \to \varphi(U)$ is a homeomorphism, $\varphi(A \cap U)$ is open in $\varphi(U)$. Similarly, $\psi(B \cap V)$ is open in $\psi(V)$. Therefore $\varphi(A \cap U) \times \psi(B \cap V)$ is open in $\varphi(U) \times \psi(V)$ with its Euclidean product subspace topology. Thus $A \times B \in \tau_{\mathcal{A}}$.
Because the sets $A \times B$ with $A \in \tau_M$ and $B \in \tau_N$ form a basis for $\tau_{\mathrm{prod}}$, every $\tau_{\mathrm{prod}}$-open set is a union of such product-basic open sets. The class $\tau_{\mathcal{A}}$ is closed under arbitrary unions: for any family $(W_i)_{i \in I}$ of $\tau_{\mathcal{A}}$-open sets and any product chart $\Phi_{\varphi,\psi}: U \times V \to \varphi(U) \times \psi(V)$, the image of $(\bigcup_{i \in I} W_i) \cap (U \times V)$ under $\Phi_{\varphi,\psi}$ is the union of the open sets $\Phi_{\varphi,\psi}(W_i \cap (U \times V))$. Hence that image is open in $\varphi(U) \times \psi(V)$. Therefore arbitrary unions of $\tau_{\mathcal{A}}$-open sets are $\tau_{\mathcal{A}}$-open, and it follows that $\tau_{\mathrm{prod}} \subseteq \tau_{\mathcal{A}}$.
[/step]
[step:Conclude that the product manifold topology is the product topology]
The previous two inclusions give
\begin{align*}
\tau_{\mathcal{A}} \subseteq \tau_{\mathrm{prod}}
\end{align*}
and
\begin{align*}
\tau_{\mathrm{prod}} \subseteq \tau_{\mathcal{A}}.
\end{align*}
Hence
\begin{align*}
\tau_{\mathcal{A}} = \tau_{\mathrm{prod}}.
\end{align*}
Therefore the topology underlying the standard smooth product manifold structure on $M \times N$ is exactly the product topology on $M \times N$.
[/step]
