Given, the cartesian product of two non-empty sets A and B (subsets of a metric space M) is sequentially compact, show that A and B are compact We know that if $A$ and $B$ are compact (assuming A and B are non-empty), then the Cartesian product $A \text{x} B$ is compact. But how do you go the other way round.
We have to show that any sequence $(a_k)$ in A and $(b_k)$ in B have subsequences that converges in A and B respectively. We are given that any subseqence of the sequence $(a_k, b_k)$ is convergent in $A \text{ x } B$. I have at loss at how to I use this information to claim that subsequences of $(a_k) \text{and} (b_k)$ are convergent in $A$ and $B$ respectively. 
Should I claim that $(a_k, b_k)$ is convergent iff each $a_k$ and $b_k$ is convergent, and be done with it?  
 A: HINT: Let $\langle a_k:k\in\Bbb N\rangle$ be a sequence in $A$. Let $b$ be any point of $B$, and consider the sequence $$\big\langle\langle a_k,b\rangle:k\in\Bbb N\big\rangle$$ in $A\times B$. The same basic idea works to handle sequences in $B$.
Added: Here’s a rough sketch that may help a bit.

I’ve drawn this as if $A$ and $B$ were $[0,1]$, because that’s easy to visualize. The red dots in $A$ (down at the bottom) are four terms of the sequence $\langle a_k:k\in\Bbb N\rangle$; the red dots in $A\times B$ (the square) are all on the line representing $A\times\{b\}$: they are the corresponding terms of the sequence $$\big\langle\langle a_k,b\rangle:k\in\Bbb N\big\rangle\tag{1}$$ in $A\times B$.
You know that every sequence in $A\times B$ has a convergent subsequence, so $(1)$ converges to some point $\langle x,y\rangle\in A\times B$. In fact, $y=b$; why? Thus, $\langle x,y\rangle=\langle x,b\rangle$ is on the same ‘line’ as the sequence $(1)$, the set $A\times\{b\}$. Now show, using the definition of the product topology, that $\langle a_k:k\in\Bbb N\rangle$ converges to $x$.
