There are given two metric spaces $(X_1, \rho_1)$ and $(X_2, \rho_2)$, we define metric on product $X_1 \times X_2$ as follows $\rho = \rho_1 + \rho_2$.
Show that $(X_1\times X_2, \rho)$ is complete iff both $(X_1, \rho_1)$ and $(X_2, \rho_2)$ are complete metric spaces.
($\Leftarrow$) Suppose that both are complete. Let $x_n = (x^{(1)}_n, x^{(2)}_n)$ be Cauchy sequence in product space. This means that for every $\epsilon > 0$ there exists $N$ such that for ever $n,m>N$ we have $\rho(x_n,x_m) < \epsilon$
which in particular implies that $\rho_1(x^{(1)}_n,x^{(1)}_m), \rho_2(x^{(2)}_n,x^{(2)}_m) < \epsilon$.
By the assumption of completeness of $X_1,X_2$ we receive that $x^{(1)}_n \rightarrow x^{(1)}$, $x^{(2)}_n \rightarrow x^{(2)}$ in $X_1,X_2$ respectively.
Now we have to show that $(x^{(1)},x^{(2)})$ is a limit in product space. But from convergence of these two we find sufficiently large $N$ ($max\{N_1,N_2\}$) such that
$\rho((x^{(1)}_n, x^{(2)}_n), (x^{(1)},x^{(2)})) = \rho_1(x^{(1)}_n,x^{(1)})+\rho_1(x^{(2)}_n,x^{(2)}) < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$ for all $n>N$.
Is it correct so far?
About the $(\Rightarrow)$. Is it fine to take $(x^{(1)}_n)$ Cauchy sequence in $X_1$ then pick constant sequence $(x^{(2)})$ in $X_2$. And then just from the completeness of product and by the fact that $(x^{(1)}_n, x^{(2)})$ is Cauchy sequence in product one can deduce that $X_1$ is also complete? Then do it again, but for $X_2$.