If $C\subseteq [0,1]$ is uncountable then $C\cap[\alpha,1]$ is uncountable

Question

I am asked to prove or disprove the following:

Let $C\subseteq[0,1]$ be uncountable, and let $A$ be the set of all values $a\in(0,1)$ such that $C\cap[a,1]$ is uncountable. Define $\alpha=\sup A$. Is $C\cap[\alpha ,1]$ also uncountable?

I haven’t been able to come up with a proof, but I have the following counter example to disprove the statement:

Fix $b\in(0,1)$, and let $C=\lbrace x\in\mathbb{R} | x\in[0,b)\rbrace\cap \lbrace x\in\mathbb{Q} | x\in[b,1]\rbrace$. Then, $C\cap[\alpha ,1]$ is countable.

I have the following questions:

$1)$ Can I prove this geberally without the use of a counterexample?

$2)$ Why is the choice for $a$ restricted to $(0,1)$?

Answer 1

1. No need to prove anything, one counterexample suffice to prove a statement is false.
2. Because for $a=1$, we have $[a, 1]=\{1\} $ countable, and for $a=0$, it's $[0,1]\cap C=C$ uncountable by hypothesis.
3. However, if we take $\inf$ instead of $\sup$, the statement becomes true. Can you prove it?

Answer 2

Take $C = [0,1]$. Then we have $A = (0,1)$, and $\sup A = 1$. But $C \cap [a,1] = \{1\}$.

Answer 3

A direct proof:

Since $\alpha = \sup A$, it follows that $C\cap(\alpha , 1]$ is countable.

Suppose $\alpha \notin C$. Then $C\cap (\alpha, 1] = C \cap [\alpha, 1]$, so RHS is countable.

Suppose $\alpha\in C$. Then $\{\alpha\} \cup (C \cap (\alpha, 1]) = (\{\alpha\} \cup C) \cap (\{\alpha\} \cup (\alpha, 1]) = C \cap [\alpha, 1]$ is countable.