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update: the argument does not seem to work
Alexey Muranov
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Thanks for all the answers and sorry about a silly question. I have also figured out that it can be proved using the usual complete metric on the usual (countable product) Hilbert cube and finite $\epsilon$-nets.

Update: I think finite $\epsilon$-nets even explicitly chosen only give countable compactness.

Alexey Muranov
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