Please note that the title is not the complete question I am asking, it just serves as a topic summary. The question is below.
I am hoping someone can review my proof. I have put a (*) next to the claim I am personally most uncertain about. Thanks in advance!
Question:
Let X have a countable basis; let A be an uncountable subset of X. Show that uncountably many points of A are limit points of A.
Proof:
Let X be a topological space with a countable basis. Let A be an uncountable subset of X.
Let A' be the set of limit points of A that are contained within A.
Suppose A' is not uncountable. Then A' is countable.
Since A is uncountable, we have B = A - A' is non-empty.
(*) Also we have that B must be uncountable since if B was countable we would have:
$$ A = B \bigcup A' $$
But then A would be the union of two countable sets, which is countable. Yet A was chosen to be uncountable. So B must be uncountable.
So we have B $\subset$ A and B consists of an uncountable number of points each of which is a member of A yet not a limit point of A. Then we for each x$_i$ $\in$ B we can find an open set U$_i$ such that x $\in$ U$_i$ and U$_i$ $\bigcap$ A = {x$_i$}. Then since U$_i$ is open it must contain some basis element B$_i$ such that x$_i$ $\in$ B$_i$ $\subseteq$ U$_i$.
Now the set of B$_i$ basis elements is an uncountable set of basis elements of X since for each unique $x_i$ we produce a unique B$_i$. Consider that for some x$_n$ and x$_z$ that their corresponding open basis elements B$_n$ and B$_z$ cannot be the same basis elements since if either were, the basis element would contain both x$_n$ and x$_z$ which is not allowed since the points of B are limit points and the open set B$_i$ must only intersect A in one single point, that is x$_i$.
Hence we have an uncountable collection of basis elements in a space that has a countable basis. A contradiction. So A must contain uncountably many points that are limit points of A.