Recall that a space has covering dimension zero if every open cover of it can be refined to a disjoint open cover.
Question 1. Does the product of two spaces of covering dimension zero have covering dimension zero?
Question 2. Does the product of infinitely many spaces of covering dimension zero have covering dimension zero?
Question 3. Are the answers different for topological spaces and locales?
I think I have a proof that works for compact spaces of covering dimension zero, using arguments similar to the one to prove the tube lemma and the Tychonoff theorem.
However, there are certainly non-compact spaces of covering dimension zero.
Best Answer
It should be well-known that a regular, T1 space has covering dimension zero, iff it is paracompact and strongly zero-dimensional. Hence, a Lindelof zero-dimensional T1 space has covering dimension zero (see Engelking, General topology 6.2.7).
In particular, the Sorgenfrey line has covering dimension zero, but its square is not normal, hence does not have covering dimension zero.
Update
It should be noted that the above definition of "covering dimension zero" is not the usual one. The usual one only requires that $X$ is Tychonoff and every finite cover of cozero sets has a finite open refinement, i.e. it is strongly zero-dimensional. Of course, my above answer is according to the OP's definition. Spaces with this property are frequently called "ultraparacompact".
According to Engelking, General topology, p.363, there also exist two strongly zero-dimensional spaces such that their product is not strongly zero-dimensional. However, these examples seem to be very difficult. (Therefore, they are not even described there.)