[proofplan]
We compare neighborhoods at an arbitrary point $(x,y) \in X \times Y$. First, every $d_p$-ball contains a product of coordinate balls, so every $d_p$-[open set](/page/Open%20Set) is open in the [product topology](/page/Product%20Topology). Second, every basic product neighborhood contains a sufficiently small $d_p$-ball, using the coordinate bounds $d_X(x,x') \le d_p((x,y),(x',y'))$ and $d_Y(y,y') \le d_p((x,y),(x',y'))$. These two local containment statements prove equality of the two topologies.
[/proofplan]
[step:Fix the two topologies and the notation for balls]
Let $\tau_X$ be the metric topology on $X$ induced by $d_X$, let $\tau_Y$ be the metric topology on $Y$ induced by $d_Y$, and let $\tau_{\mathrm{prod}}$ be the product topology on $X \times Y$ generated by sets of the form $U \times V$ with $U \in \tau_X$ and $V \in \tau_Y$.
Let $\tau_p$ denote the topology on $X \times Y$ induced by the [product metric](/page/Product%20Metric) $d_p$ specified in the theorem statement.
For $x \in X$ and $r>0$, define the $d_X$-ball
\begin{align*}
B_X(x,r)=\{x' \in X : d_X(x,x')<r\}.
\end{align*}
For $y \in Y$ and $r>0$, define the $d_Y$-ball
\begin{align*}
B_Y(y,r)=\{y' \in Y : d_Y(y,y')<r\}.
\end{align*}
For $(x,y) \in X \times Y$ and $r>0$, define the $d_p$-ball
\begin{align*}
B_p((x,y),r)=\{(x',y') \in X \times Y : d_p((x,y),(x',y'))<r\}.
\end{align*}
[/step]
[step:Place a product of coordinate balls inside each $d_p$-ball]
Fix $(x,y) \in X \times Y$ and $\varepsilon>0$. We claim that
\begin{align*}
B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2) \subset B_p((x,y),\varepsilon).
\end{align*}
Let $(x',y') \in B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2)$. Then
\begin{align*}
d_X(x,x')<\varepsilon/2
\end{align*}
and
\begin{align*}
d_Y(y,y')<\varepsilon/2.
\end{align*}
If $1 \le p < \infty$, then
\begin{align*}
d_p((x,y),(x',y'))=\left(d_X(x,x')^p+d_Y(y,y')^p\right)^{1/p}.
\end{align*}
Using the two coordinate bounds gives
\begin{align*}
d_p((x,y),(x',y'))<\left((\varepsilon/2)^p+(\varepsilon/2)^p\right)^{1/p}.
\end{align*}
The right-hand side equals $2^{1/p}\varepsilon/2$. Since $p \ge 1$, we have $2^{1/p} \le 2$, and therefore
\begin{align*}
d_p((x,y),(x',y'))<\varepsilon.
\end{align*}
If $p=\infty$, then
\begin{align*}
d_\infty((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')\}<\varepsilon/2<\varepsilon.
\end{align*}
Thus the claimed containment holds for every $1 \le p \le \infty$.
It follows that every $\tau_p$-open set is $\tau_{\mathrm{prod}}$-open: if $W \in \tau_p$ and $(x,y) \in W$, choose $\varepsilon>0$ with $B_p((x,y),\varepsilon)\subset W$. The product neighborhood $B_X(x,\varepsilon/2)\times B_Y(y,\varepsilon/2)$ lies in $W$ by the containment above, so $W$ is open in the product topology.
[guided]
Fix a point $(x,y) \in X \times Y$ and a radius $\varepsilon>0$. Recall that
\begin{align*}
B_X(x,r)=\{x' \in X : d_X(x,x')<r\}
\end{align*}
denotes the open ball in $(X,d_X)$, that
\begin{align*}
B_Y(y,r)=\{y' \in Y : d_Y(y,y')<r\}
\end{align*}
denotes the open ball in $(Y,d_Y)$, and that
\begin{align*}
B_p((x,y),r)=\{(x',y') \in X \times Y : d_p((x,y),(x',y'))<r\}
\end{align*}
denotes the open ball in the product metric. To show that the $d_p$-topology is no finer than the product topology, we must show that a $d_p$-ball contains a basic product neighborhood of its center. The natural candidate is a product of two small coordinate balls:
\begin{align*}
B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2).
\end{align*}
Take an arbitrary point $(x',y')$ in this product. By membership in the first coordinate ball,
\begin{align*}
d_X(x,x')<\varepsilon/2.
\end{align*}
By membership in the second coordinate ball,
\begin{align*}
d_Y(y,y')<\varepsilon/2.
\end{align*}
Now we verify that $(x',y')$ is inside the $d_p$-ball of radius $\varepsilon$. If $1 \le p < \infty$, the definition of the product metric gives
\begin{align*}
d_p((x,y),(x',y'))=\left(d_X(x,x')^p+d_Y(y,y')^p\right)^{1/p}.
\end{align*}
Substituting the two strict coordinate inequalities gives
\begin{align*}
d_p((x,y),(x',y'))<\left((\varepsilon/2)^p+(\varepsilon/2)^p\right)^{1/p}.
\end{align*}
The expression on the right simplifies to $2^{1/p}\varepsilon/2$. Since $p \ge 1$, the inequality $2^{1/p}\le 2$ holds, so
\begin{align*}
2^{1/p}\varepsilon/2 \le \varepsilon.
\end{align*}
Hence
\begin{align*}
d_p((x,y),(x',y'))<\varepsilon.
\end{align*}
For the endpoint $p=\infty$, the metric is the maximum of the coordinate distances. The same coordinate bounds give
\begin{align*}
d_\infty((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')\}<\varepsilon/2<\varepsilon.
\end{align*}
Therefore
\begin{align*}
B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2) \subset B_p((x,y),\varepsilon)
\end{align*}
for every $1 \le p \le \infty$.
This containment converts metric openness into product openness. Let $W$ be open in $\tau_p$, and let $(x,y)\in W$. By definition of the metric topology, there exists $\varepsilon>0$ such that
\begin{align*}
B_p((x,y),\varepsilon)\subset W.
\end{align*}
The containment just proved supplies a basic product neighborhood
\begin{align*}
B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2)
\end{align*}
with
\begin{align*}
(x,y)\in B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2)\subset W.
\end{align*}
Thus every point of $W$ has a basic product neighborhood contained in $W$, which is precisely the criterion that $W$ be open in the product topology.
[/guided]
[/step]
[step:Place a $d_p$-ball inside each basic product neighborhood]
Let $U \in \tau_X$ and $V \in \tau_Y$, and fix $(x,y) \in U \times V$. Since $U$ is open in the metric topology induced by $d_X$, there exists $r_X>0$ such that
\begin{align*}
B_X(x,r_X)\subset U.
\end{align*}
Since $V$ is open in the metric topology induced by $d_Y$, there exists $r_Y>0$ such that
\begin{align*}
B_Y(y,r_Y)\subset V.
\end{align*}
Define
\begin{align*}
r=\min\{r_X,r_Y\}.
\end{align*}
Then $r>0$.
We claim that
\begin{align*}
B_p((x,y),r)\subset U \times V.
\end{align*}
Let $(x',y') \in B_p((x,y),r)$. If $1 \le p < \infty$, then
\begin{align*}
d_X(x,x') \le \left(d_X(x,x')^p+d_Y(y,y')^p\right)^{1/p}=d_p((x,y),(x',y'))<r.
\end{align*}
Also,
\begin{align*}
d_Y(y,y') \le \left(d_X(x,x')^p+d_Y(y,y')^p\right)^{1/p}=d_p((x,y),(x',y'))<r.
\end{align*}
If $p=\infty$, then
\begin{align*}
d_X(x,x')\le d_\infty((x,y),(x',y'))<r
\end{align*}
and
\begin{align*}
d_Y(y,y')\le d_\infty((x,y),(x',y'))<r.
\end{align*}
In all cases, $d_X(x,x')<r\le r_X$ and $d_Y(y,y')<r\le r_Y$, so $x'\in U$ and $y'\in V$. Hence $(x',y')\in U\times V$.
Therefore every basic product neighborhood contains a $d_p$-ball around each of its points. Since arbitrary product-open sets are unions of such basic product neighborhoods, every $\tau_{\mathrm{prod}}$-open set is $\tau_p$-open.
[/step]
[step:Conclude that the two topologies are equal]
The previous two steps prove the two inclusions
\begin{align*}
\tau_p \subset \tau_{\mathrm{prod}}
\end{align*}
and
\begin{align*}
\tau_{\mathrm{prod}} \subset \tau_p.
\end{align*}
Therefore
\begin{align*}
\tau_p=\tau_{\mathrm{prod}}.
\end{align*}
Thus the topology induced by $d_p$ on $X\times Y$ is exactly the product topology induced by the metric topologies on $X$ and $Y$.
[/step]
