Intersection of uncountable sets. Let $C \subseteq  [0,1]$ be uncountable. Let $A$ be the set of all $a \in (0, 1) $ such that $ C \cap [a,1] $ is uncountable and set $\alpha = \sup A$. Is $C \cap [\alpha, 1]$ an uncountable set?
I know that $a$ must exist, but I'm not sure how to go about proving/disproving this statement. 
 A: Take $C=[0,1]$. Then $A= (0,1)$, hence $\sup(A)=1$. Then $C \cap [\sup(A),1]= \left\{ 1 \right \}$, which is countable.
A: Suppose $C\subseteq [0,1]$ is uncountable. Let $A = \{a\in (0,1)\mid C\cap[a,1] \text{ is uncountable} \}$, and $\alpha = \sup A$. 
Then $C\cap [\alpha,1]$ is countable.
First, note that $A$ is nonempty: for $n\in\Bbb N_+$ let $C_n = C\cap [\frac 1 n, 1]$. Some $C_n$ must be uncountable, otherwise $C= \bigcup_n C_n$ is a countable union of countable sets and therefore countable. So for some $n$, $1/n\in A$.
Clearly $0 < \alpha \le 1$.
If $\alpha =1 $ then of course the claim is true, so suppose $\alpha < 1$. Let $(b_n)$ be a decreasing sequence in $(\alpha, 1)$ with $\alpha = \inf_n b_n$. By definition of $A$ and $\alpha$, for every $n$, $C\cap[b_n,1]$ is countable, for otherwise $b_n\in A$ and $b_n \le \alpha$. Thus 
$$\begin{align}
C\cap [\alpha,1] &= C\cap \bigcup_n [b_n, 1] \\
&= \bigcup_n (C\cap [b_n, 1])
\end{align}$$
is a countable union of countable sets, so it's countable.
