Yes: at least one factor topology must be indiscrete. Suppose instead that both have a nonempty proper open set, say and . Then
is open in the product topology. Both coordinate projections of are the full factors, because are nonempty. If were a rectangle , this would force , , hence . But choosing and gives , a contradiction. Thus both factors cannot have nontrivial open sets. This is rectangularity of every product-open set.

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