[proofplan]
The proof uses only the defining formulas for the product metrics. First we record the coordinate inequalities $d_X \le d_p$ and $d_Y \le d_p$, which show that product-Cauchy sequences have coordinatewise Cauchy sequences. For one direction, completeness of the factors gives coordinate limits, and the metric formula then gives convergence in the product. For the converse, nonemptiness lets us embed each factor into the product by fixing a basepoint in the other factor, so completeness of the product forces completeness of each factor.
[/proofplan]
[step:Extract the coordinate inequalities from the product metric]
For all $(x_1,y_1),(x_2,y_2)\in X\times Y$, we have
\begin{align*}
d_X(x_1,x_2)\le d_p((x_1,y_1),(x_2,y_2))
\end{align*}
and
\begin{align*}
d_Y(y_1,y_2)\le d_p((x_1,y_1),(x_2,y_2)).
\end{align*}
Indeed, if $1\le p<\infty$, then $d_X(x_1,x_2)^p$ and $d_Y(y_1,y_2)^p$ are nonnegative [real numbers](/page/Real%20Numbers), so
\begin{align*}
d_X(x_1,x_2)^p\le d_X(x_1,x_2)^p+d_Y(y_1,y_2)^p
\end{align*}
and
\begin{align*}
d_Y(y_1,y_2)^p\le d_X(x_1,x_2)^p+d_Y(y_1,y_2)^p.
\end{align*}
Taking $p$th roots gives both inequalities. If $p=\infty$, both inequalities follow directly from the definition of the maximum:
\begin{align*}
d_X(x_1,x_2)\le \max\{d_X(x_1,x_2),d_Y(y_1,y_2)\}
\end{align*}
and
\begin{align*}
d_Y(y_1,y_2)\le \max\{d_X(x_1,x_2),d_Y(y_1,y_2)\}.
\end{align*}
[/step]
[step:Prove product completeness from completeness of both factors]
Assume that $(X,d_X)$ and $(Y,d_Y)$ are complete. Let $z:\mathbb N\to X\times Y$ be a $d_p$-[Cauchy sequence](/page/Cauchy%20Sequence), and write $z(n)=(x_n,y_n)$ for the coordinate sequences $x:\mathbb N\to X$ and $y:\mathbb N\to Y$.
By the coordinate inequalities, for all $m,n\in\mathbb N$,
\begin{align*}
d_X(x_m,x_n)\le d_p(z(m),z(n))
\end{align*}
and
\begin{align*}
d_Y(y_m,y_n)\le d_p(z(m),z(n)).
\end{align*}
Hence $x$ is a Cauchy sequence in $(X,d_X)$ and $y$ is a Cauchy sequence in $(Y,d_Y)$. Completeness of the factors gives points $x_\ast\in X$ and $y_\ast\in Y$ such that $x_n\to x_\ast$ in $(X,d_X)$ and $y_n\to y_\ast$ in $(Y,d_Y)$.
It remains to show that $z(n)\to (x_\ast,y_\ast)$ in $(X\times Y,d_p)$. Let $\varepsilon>0$. If $1\le p<\infty$, choose $N_X\in\mathbb N$ such that
\begin{align*}
d_X(x_n,x_\ast)<\frac{\varepsilon}{2^{1/p}}
\end{align*}
for all $n\ge N_X$, and choose $N_Y\in\mathbb N$ such that
\begin{align*}
d_Y(y_n,y_\ast)<\frac{\varepsilon}{2^{1/p}}
\end{align*}
for all $n\ge N_Y$. For $n\ge N:=\max\{N_X,N_Y\}$, the defining formula gives
\begin{align*}
d_p(z(n),(x_\ast,y_\ast))^p=d_X(x_n,x_\ast)^p+d_Y(y_n,y_\ast)^p<\frac{\varepsilon^p}{2}+\frac{\varepsilon^p}{2}=\varepsilon^p.
\end{align*}
Taking $p$th roots gives $d_p(z(n),(x_\ast,y_\ast))<\varepsilon$.
If $p=\infty$, choose $N_X\in\mathbb N$ such that $d_X(x_n,x_\ast)<\varepsilon$ for all $n\ge N_X$, and choose $N_Y\in\mathbb N$ such that $d_Y(y_n,y_\ast)<\varepsilon$ for all $n\ge N_Y$. For $n\ge N:=\max\{N_X,N_Y\}$,
\begin{align*}
d_\infty(z(n),(x_\ast,y_\ast))=\max\{d_X(x_n,x_\ast),d_Y(y_n,y_\ast)\}<\varepsilon.
\end{align*}
Thus $z(n)\to (x_\ast,y_\ast)$, so every $d_p$-Cauchy sequence in $X\times Y$ converges. Therefore $(X\times Y,d_p)$ is complete.
[guided]
Assume that both factor spaces are complete. To prove that $(X\times Y,d_p)$ is complete, we must start with an arbitrary Cauchy sequence in the product and prove that it converges in the [product metric](/page/Product%20Metric).
Let $z:\mathbb N\to X\times Y$ be a $d_p$-Cauchy sequence. Since every element of $X\times Y$ has two coordinates, write $z(n)=(x_n,y_n)$, where $x:\mathbb N\to X$ is the first-coordinate sequence and $y:\mathbb N\to Y$ is the second-coordinate sequence.
The first point is that Cauchy behavior in the product controls Cauchy behavior in each coordinate. From the coordinate inequalities proved above, for all $m,n\in\mathbb N$,
\begin{align*}
d_X(x_m,x_n)\le d_p(z(m),z(n))
\end{align*}
and
\begin{align*}
d_Y(y_m,y_n)\le d_p(z(m),z(n)).
\end{align*}
Because $z$ is $d_p$-Cauchy, for every $\varepsilon>0$ there is $N\in\mathbb N$ such that $d_p(z(m),z(n))<\varepsilon$ whenever $m,n\ge N$. The two displayed inequalities then give $d_X(x_m,x_n)<\varepsilon$ and $d_Y(y_m,y_n)<\varepsilon$ for all $m,n\ge N$. Hence $x$ is Cauchy in $(X,d_X)$ and $y$ is Cauchy in $(Y,d_Y)$.
Now the completeness hypotheses are used exactly once: since $(X,d_X)$ is complete, there exists $x_\ast\in X$ such that $x_n\to x_\ast$ in $d_X$; since $(Y,d_Y)$ is complete, there exists $y_\ast\in Y$ such that $y_n\to y_\ast$ in $d_Y$. The only remaining question is whether coordinate convergence implies product convergence for this particular product metric. We verify this directly from the formula.
Let $\varepsilon>0$. First suppose $1\le p<\infty$. Since $x_n\to x_\ast$, there is $N_X\in\mathbb N$ such that
\begin{align*}
d_X(x_n,x_\ast)<\frac{\varepsilon}{2^{1/p}}
\end{align*}
for all $n\ge N_X$. Since $y_n\to y_\ast$, there is $N_Y\in\mathbb N$ such that
\begin{align*}
d_Y(y_n,y_\ast)<\frac{\varepsilon}{2^{1/p}}
\end{align*}
for all $n\ge N_Y$. For $n\ge N:=\max\{N_X,N_Y\}$, both estimates hold simultaneously. Substituting them into the finite-$p$ product formula gives
\begin{align*}
d_p(z(n),(x_\ast,y_\ast))^p=d_X(x_n,x_\ast)^p+d_Y(y_n,y_\ast)^p<\frac{\varepsilon^p}{2}+\frac{\varepsilon^p}{2}=\varepsilon^p.
\end{align*}
Because both sides are nonnegative, taking $p$th roots yields $d_p(z(n),(x_\ast,y_\ast))<\varepsilon$.
Now suppose $p=\infty$. The product distance is controlled by the larger coordinate distance, so we ask that each coordinate distance be smaller than $\varepsilon$. Choose $N_X\in\mathbb N$ such that $d_X(x_n,x_\ast)<\varepsilon$ for $n\ge N_X$, and choose $N_Y\in\mathbb N$ such that $d_Y(y_n,y_\ast)<\varepsilon$ for $n\ge N_Y$. Then for every $n\ge N:=\max\{N_X,N_Y\}$,
\begin{align*}
d_\infty(z(n),(x_\ast,y_\ast))=\max\{d_X(x_n,x_\ast),d_Y(y_n,y_\ast)\}<\varepsilon.
\end{align*}
Thus $z(n)\to (x_\ast,y_\ast)$ in the product metric. Since the original Cauchy sequence $z$ was arbitrary, every Cauchy sequence in $(X\times Y,d_p)$ converges, and therefore $(X\times Y,d_p)$ is complete.
[/guided]
[/step]
[step:Recover completeness of each factor from product completeness]
Assume that $(X\times Y,d_p)$ is complete. Since $Y$ is nonempty, choose $y_0\in Y$. Let $x:\mathbb N\to X$ be a Cauchy sequence in $(X,d_X)$, and define $z_X:\mathbb N\to X\times Y$ by
\begin{align*}
z_X(n)=(x_n,y_0).
\end{align*}
For all $m,n\in\mathbb N$, if $1\le p<\infty$, then
\begin{align*}
d_p(z_X(m),z_X(n))=\bigl(d_X(x_m,x_n)^p+d_Y(y_0,y_0)^p\bigr)^{1/p}=d_X(x_m,x_n).
\end{align*}
If $p=\infty$, then
\begin{align*}
d_\infty(z_X(m),z_X(n))=\max\{d_X(x_m,x_n),d_Y(y_0,y_0)\}=d_X(x_m,x_n).
\end{align*}
Thus $z_X$ is a Cauchy sequence in $(X\times Y,d_p)$. By product completeness, there exists $(x_\ast,y_\ast)\in X\times Y$ such that $z_X(n)\to (x_\ast,y_\ast)$ in $d_p$. The coordinate inequality gives
\begin{align*}
d_X(x_n,x_\ast)\le d_p(z_X(n),(x_\ast,y_\ast)).
\end{align*}
Hence $x_n\to x_\ast$ in $(X,d_X)$. Since $x$ was an arbitrary Cauchy sequence in $X$, the space $(X,d_X)$ is complete.
Since $X$ is nonempty, choose $x_0\in X$. Let $y:\mathbb N\to Y$ be a Cauchy sequence in $(Y,d_Y)$, and define $z_Y:\mathbb N\to X\times Y$ by
\begin{align*}
z_Y(n)=(x_0,y_n).
\end{align*}
The same computation, with the roles of $X$ and $Y$ exchanged, gives
\begin{align*}
d_p(z_Y(m),z_Y(n))=d_Y(y_m,y_n)
\end{align*}
for all $m,n\in\mathbb N$, with the displayed identity interpreted by the finite-$p$ formula when $1\le p<\infty$ and by the maximum formula when $p=\infty$. Thus $z_Y$ is Cauchy in the complete product, so $z_Y(n)\to (x_\ast',y_\ast')$ for some $(x_\ast',y_\ast')\in X\times Y$. The coordinate inequality gives
\begin{align*}
d_Y(y_n,y_\ast')\le d_p(z_Y(n),(x_\ast',y_\ast')).
\end{align*}
Therefore $y_n\to y_\ast'$ in $(Y,d_Y)$. Since $y$ was arbitrary, $(Y,d_Y)$ is complete.
[/step]
[step:Combine the two implications]
The previous step shows that product completeness implies completeness of both factors. The product-completeness step shows that completeness of both factors implies completeness of the product. Therefore $(X\times Y,d_p)$ is complete if and only if both $(X,d_X)$ and $(Y,d_Y)$ are complete.
[/step]
