Yes, "A \\setminus B" must be uncountable.
Proof. Assume "A \\setminus B" is countable. Then, we can express "A" as "(A \\setminus B) \\bigcup B". Since the union of two countable sets is countable, "A" must be countable. This is a contradiction, and so "A \\setminus B" is uncountable.
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