How to prove that the given set is not uncountable? I was trying to solve the question given in my assignment on metric spaces.
Let $S$ be a subset of $R$. Let $C$ be the set of points $x$ in $R$ with the property that $S\cap (x-\delta,x+\delta )$ is uncountable for every $\delta > 0.$ Prove that $S - C$ is finite or countable. 
I started like this:
Let $x$ $\in$ $S-C$ $=>$ $S\cap (x-\delta,x+\delta )$ is countable for some $\delta>0$.
Also, $(S-C)\cap(x-\delta,x+\delta )\subset S\cap (x-\delta,x+\delta )$.
So, $(S-C)\cap(x-\delta,x+\delta )$ is countable.
Hence, for every $x$ $ \in$ $S-C$ , $\exists$ a $\delta>0$ such that $(S-C)\cap(x-\delta,x+\delta )$ is countable.
But after that I couldn't advance. Any help would be appreciated.
Thanks.
 A: In your question you got to this step:

Hence, for every $x$ $ \in$ $S-C$ , $\exists$ a $\delta>0$ such that $(S-C)\cap(x-\delta,x+\delta )$ is countable.

Note that the collection of all the sets $(S-C)\cap(x-\delta,x+\delta )$ (choose one such set for each $x)$ gives a covering of $S-C$ by open intervals. Now use the fact (or prove it if you don't yet have it available) that given any collection of open intervals, there exists an at most countable subcollection of the original collection whose union is the same as the union of the original collection.
A: If $|S-C|$ is uncountable, then there exists a point $x\in S-C$ s.t $N_\delta (x)\cap S$ is uncountable for all $\delta$. 
(observe that given an uncountable set $A$, there exists a point $a$ whose any $\delta$ nbd contains uncountably many point of $A$.... Now take $A=S-C$
proof of the observation suppose for any $a\in A$ there exists a $\delta_a$ s.t $N_{\delta_a}(A)\cap A$ is countable. for a fix $a\in A$ define $A' = A- N_{\delta_a}(a)$. Then $A'$ is uncountable. And we can repeat the same operation for a point $b$ in $A'$ s.t $N_{\delta_a}(a)\cap N_{\delta_b}(b) = \phi$. Now if we keep on repeating this process only countably many time, because any open set in $\mathbb R$ is countable union of open sets. And thus we are only eliminating countably many point. So still we are left with some points. And for those point then any nbd contains uncountably many point of $A$. NOTE: if you stay a bit more careful, then you can end up with a perfect set in $A$).
In this case $x\in C$. Which is a contradiction.
A: For any $x\in S\setminus C$, let $a_x,b_x$ be defined so that:
$$a_x=\inf\{a\mid S\cap (a,x) \text{ is countable or finite}\}\\
b_x=\sup\{b\mid S\cap (x,b)\text{ is countable or finite}\}$$
There is always such $a_x,b_x$ because $x\notin C$ so $(x-\delta,x+\delta)$ is finite or countable for some $\delta$, and we can show that $(a_x,b_x)\cap S$ is countable or finite, since it is the countable union of $S\cap\left(a_x+\frac{1}{n},b_x-\frac{1}{n}\right)$ each of which must be countable.
The values $a_x,b_x$ can be $-\infty$ or $+\infty$, respectively.
Now, if $x,y\in S\setminus C$ and $y\in (a_x,b_x)$ then $(a_y,b_y)=(a_x,b_x)$.
So we have an equivalence relation on $S\setminus C$ defined as $x\sim y$ if and only if $(a_x,b_x)=(a_y,b_y)$. Modulo this equivalence relation, there is only countably many classes, because $x\sim y$, and the interals $(a_x,b_x)$ and $(a_y,b_y)$ are disjoint of the are not equal.
But then:
$$S\setminus C \subseteq S\cap \bigcup_{i=1}^{\infty}(a_{x_i},b_{x_i})=\bigcup_{i=1}^{\infty} S\cap (a_{x_i},b_{x_i})$$ is countable or finite.

A more direct way to write this same proof is to use that $\mathbb R$ has a countable basis. This also lets you generalize the theorem.
For each $x\in S\setminus C$, let $\delta_x>0$ be chosen so that $I_x=(x-\delta_x,x+\delta_x)$ has the property that $I_x\cap S$ is finite or countable.
Then let $U=\bigcup_{x\in S\setminus C} I_x$. 
We see immediately that $S\setminus C\subseteq U$, and that $U$ is open, since it is a union of open sets.
Let $U_1,U_2,U_3,\dots$ be a countable basis for $\mathbb R$. Then pick the countable subset $V_1,V_2,\dots$ such that $V_i\subseteq I_x$ for some $x\in S\setminus C$.  Show $U=\bigcup V_i$.
Now, since $V_i\subseteq I_x$ for some $x$, $V_i\cap S\subseteq I_x\cap S$ which is finite or countable.
So $S\cap U=S\cap\bigcup V_i= \bigcup S\cap V_i$ is finite or countable.
But $S\setminus C\subseteq S\cap U$. So $S\setminus C$ is also countable or finite.
Generalization: Let $X$ be a topological space with a countable basis. Let $S\subseteq X$ and let $C$ be the set of elements of $x\in X$ such that $U\cap S$ is uncountable for every open set $U$ of $X$ containing $x$. Then $S\setminus C$ is countable or finite.
