Can Tychonoffs theorem for a countable number of spaces be proven with ZF plus the axiom of (countable) dependent choice?

Question

It can be proven without any form of infinite choice that the product of two compact spaces (and thus any finite product) is compact, while on the other hand, it is well known that the general form of Tychonoffs theorem implies the axiom of choice.

So a general formulation of this question would be: Is there a proof of Tychonoffs theorem where the use of AC/Zorn's lemma can be reduced to using the axiom of dependent choice up to the ordinal that indexes the product? With the case of countable products being of particular importance.

Answer 1

From Herrlich's "The Axiom of Choice", Proposition 4.72 reads as follows:

Each of the following conditions implies the subsequent ones:

1. $\sf DC$.
2. Countable products of compact spaces are compact.
3. $\sf CC$.

He goes on to remark that (2) is provable from countable choice + the Boolean Ideal theorem, and therefore the implication from (1) to (2) is not reversible in $\sf ZF$. To my understanding, the question of whether (3) implies (2) is open.

(In Compactness in countable Tychonoff products and choice the authors say the theorem is due to Pincus, who did not publish it, and it appears in a paper by Wright.)