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Question is in the title.

According to this, every discrete subspace of $[0,1]$ must be countable. But what about its closure?

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Consider the middle-thirds Cantor set, and take the midpoint of every interval that is removed during its construction.

These midpoints evidently form a discrete set.

However, every point in the Cantor set is the limit of a sequence of midpoints, and the Cantor set is uncountable.

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  • $\begingroup$ In fact, every closed nowhere dense subset of $[0,1]$ is the closure of a discrete set. $\endgroup$ – Henno Brandsma Jun 25 '17 at 21:27

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