In [Ross & Stone], it was shown that the product of continuum many separable spaces is separable while the product of more than continuum many separable spaces is not separable. My goal here is to write down a proof here as a record for future reference. While the product of continuum many separable factors is separable, it can never be hereditarily separable. Any product space with uncountably many factors, each of which has at least two points, has a subspace that is not separable. The non-separable subspace shown here is the product. The product of continuum (but uncountably) many separable spaces is an example of a space with the countable chain condition (ccc) that is not separable.
A. The Product of Continuum Many Separable Spaces
Suppose that is a separable space for each . Then the product space is separable.
Proof. The proof uses the fact that the real line has a countable base. Let be the set of all natural numbers. For each , let be a countable dense subset of . Let be the collection of all open intervals in with rational endpoints.
Let be a pairwise disjoint finite collection of . Fix a function . For this combination, we define a point . For each open interval , for all points . For all other , define . Since there are only countably many combinations, this algorithm defines a countable subspace of the product space . It follows that is dense in .
If there are many factors where continuum, just use a subspace of the real line with cardinality as an index set. Then choose a countable base of the index set and apply the same proof.
B. The Product of Greater Than Continuum Many Separable Spaces
Suppose that for each , is a separable space with at least two points. If the product space is separable, then . Thus if the cardinality of the index set is greater than continuum, then the product space can never be separable.
Proof. For each , choose two disjoint nonempty open sets in . This is possible since has at least two points and is a Hausdorff space. Let be a countable dense subset of the product space . For each , define . Each is nonempty because it is the intersection of an open set with a dense set. For , . So here is a one-to-one mapping of into the power set of the countable set . Thus .
C. The Product Space With Uncountably Many Factors Is Not Hereditarily Separable
Proof. Suppose that we have a product space where is uncountable and each factor has at least two points. Given a point , the product is the following subspace of the product space :
for only countably many .
Let be a countable subset of . Since each point disagrees with at only countably many points, choose one such that each . Then consider the open set where is open in such that and . None of the belongs to .
The product of continuum (but uncountably) many separable spaces is an example of “ccc and not separable”.
[Ross & Stone] Ross, K. A., Stone, A. H.  Product of Separable Spaces, The American Mathematical Monthly, Vol. 71, pp. 398-403.