Consider any pair of sequences $(a_n) \subset A$ and $(b_n) \subset B.$ The sequence $((a_n),(b_n)) \subset A \times B$ is contained in $A \times B.$ Hence, there is a subsequence $(a_{f(n)}, b_{g(n)}) \subset A \times B$ whose limit $(a',b') \in A \times B.$ It follows that $(a_{f(n)}) \subset (a_n) \subset A$ and $(b_{g(n)}) \subset (g_n) \subset B,$ that is, $(a_{f(n)})$ is a subsequence of the sequence $(a_n),$ which is contained in $A,$ and $(b_{g(n)})$ is a subsequence of the sequence $(b_n),$ which is contained in $B.$ Moreover, it also follows from $(a',b') \in A \times B$ that $a' \in A$ and $b' \in B.$ Therefore, the subsequences $(a_{f(n)})$ of $(a_n) \subset A$ and $(b_{g(n)})$ of $(b_n) \subset B$ have their limits $a'$ and $b',$ respectively, in $A$ and $B,$ respectively. Hence, $A$ and $B$ are compact.
Is there something missing in this proof?
Definition of compactness in the textbook:
"A subset $A$ of a metric space $M$ is (sequentially) compact if every sequence $(a_n)$ in $A$ has a subsequence $(a_{n_k})$ that converges to a limit in $A.$
This is the only definition of compactness mentioned in the textbook.
Best Answer
There's an assumption missing: that neither $A$ nor $B$ is empty. Otherwise, the proof fails (and the statment is false, by the way).
That argument works for metric spaces. For general topological spaces it doesn't because, in general, it is not true that a topological space is compact if and only if every sequence has a convergent subsequence.