I'm pretty sure I got this one right, but I'd still like to check, as always.
Let X have a countable basis, and let A be an uncountable subset of X. Show that uncountably many points of A are limit points of A.
The Attempt at a Solution
Now, since X is second-countable, A must be second-countable, too. Hence, there exists a countable subset B of A which is dense in A, i.e. Cl(B) = A. Now, Cl(B) = B U B', where B' is the set of all limit points of B. Since B is countable it follows that B' must be uncountable, and B' is exactly the set of all limit points of A, too.