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.