We know that a product of two (or finitely many) compact topological spaces is compact. And we also know that in a metric space, compactness is equivalent to sequential compactness. So a product of two sequentially compact metric spaces is sequentially compact.
My question is this:
Let $(X, d_X)$, $(Y,d_Y)$ be two sequentially compact metric spaces, and let $\{x_n \times y_n \}$ be an arbitrary sequence in the product $X \times Y$. This implies that $\{x_n\}$ is a sequence in $X$ and $\{y_n\}$ is a sequence in $Y$.
Now since $X$ is sequentially compact, there exists a subsequence $\{x_{k_n}\}$, say, of $\{x_n\}$ which is convergent; similarly there is a subsequence $\{y_{r_n}\}$ of $\{y_n\}$ which is convergent, $\{k_n\}$ and $\{r_n\}$ being strictly increasing sequences of natural numbers.
If $k_n = r_n$ for infinitely many $n$, then we can explicitly determine a convergent subsequence of the sequence $\{x_n \times y_n \}$.
How do we explicitly determine a convergent subsequence of the sequence $\{x_n \times y_n \}$ --- whose existence is gauranteed by the sequential compactness of the product $X \times Y$ --- if $k_n = r_n$ for at most finitely many $n$?