Here is Prob. 3, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
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 $A$.
This is my attempt:
By way of contradiction, assume that $D$= {$x\in{A}$. $x$ is limit point of $A$} is countable. then the set which contains all points outside $A$ IS UNCOUNTABLE. let $a$ outside $A$ then there exists basic open set $B_a$ contains $a$ such that $B_a\cap A={a}$. we can conclude there exists one-to-one function from all points outside $A$ to elements of countable basis which impossible.
Please check my solution because I feel i made a mistake.
Best Answer
You essentially seem to be having the right idea, but the proof is very unclear.
Start by picking a countable base $\{B_n: n \in \mathbb{N}\}$ of $X$.
For every $x \in A\setminus A'$ ($A' =$ the set of limit points of $A$, that you call $D$) we can pick a basic open set $B_{n(x)}$ such that $B_{n(x)} \cap A = \{x\}$. If $x, y$ in $A \setminus A'$, then if $n(x) = n(y)$, we have that $\{x\} = B_{n(x)} \cap A = B_{n(y)} \cap A = \{y\}$ so $x = y$. Hence $x \rightarrow n(x)$ is 1-1 from $A \setminus A'$ to $\mathbb{N}$, so the former set is countable.
This means that $A \cap A'$ is uncountable, as $A = (A \cap A') \cup (A \setminus A')$ is uncountable, and not a union of two countable sets; it even has the same size as $A$.