Page "Tychonoff's theorem" Paragraph 6

It is almost trivial to prove that the product of two sequentially compact spaces is sequentially compact — one passes to a subsequence for the first component and then a subsubsequence for the second component.

An only slightly more elaborate " diagonalization " argument establishes the sequential compactness of a countable product of sequentially compact spaces.

However, the product of continuum many copies of the closed unit interval fails to be sequentially compact.