Prove that an uncountable set has at least one accumulation point 
$ \text { Let A be} \text{ an uncountable set of numbers. } \\ \text { (a) Show that } A \text { has a (finite) limit point. } \\\text { (b) Is it possible for } A \text { to have at most a finite number of limit points?} $

The Hint the textbook gives is: 
(a) Let $A_{n}=[n, n+1] \cap A$ for $n,$ an integer. If each $A_{n}$ is countable, then $\cup_{n \in \mathbb{Z}} A_{n}=$
$A$ is countable. Thus some interval $[n, n+1]$ must contain an uncountable number
of points.
I have no idea how to use this hint to go about solving this problem using introductory real analysis topics.
 A: (a) We have $A = \bigcup_{n\in\mathbb Z}([n,n+1]\cap A)$. If each of these sets was finite, then $A$ would be countable. So there is some $n$ such that $[n,n+1]\cap A$ is infinite. Now, choose a sequence $(a_k)$ with points in this set. This sequence is bounded and thus has a limit point (which is then also a limit point of $A$).
(b) By a similar reasoning as above we can show that there is $n$ such that $A_n = [n,n+1]\cap A$ is uncountable. Assume that $A_n$ has only one limit point $a$. Let $k\in\mathbb N$ and consider $A_n\setminus (a-\frac 1k,a+\frac 1k)$. There can only be finitely many points in this set since otherwise the set would have a limit point which is not $a$. So, $A_n\subset\{a\}\cup\bigcup_k(A_n\setminus (a-\frac 1k,a+\frac 1k))$ is countable, a contradiction. I am sure that you can generalize this reasoning to the case where $A_n$ has only finitely many limit points. 
A: Let $A$ be an uncountable subset of $\Bbb R.$ 
$\Bbb Q$ is countable so $\Bbb Q\times \Bbb Q$ is countable so any subset of $\{(q_1,q_2):q_1,q_2\in \Bbb Q\}$ is a countable set of  intervals. 
Let $x\in B$ iff $[\;x\in A$ and $x$ is $not$ a limit point of $A\;].$
For each $x\in B$ let $q_x^-$ and $q_x^+$ be in $\Bbb Q$ such that $q_x^-<x<q_x^+$ and $A\cap (q_x^-,q_x^+)=\{x\}.$
Let $C=\{(q_x^-,q_x^+):x\in B\}.$ Now $C$ is a countable set of intervals and each member of $C$ contains exactly one member of $B,$ while every $x\in B$ belongs to exactly one member of $C.$
So $B$ is countable.
So $A\setminus B$ is uncountable. By definition of $B,$ every member of $A\setminus B$ is a limit point of $A.$
