If $C \subseteq [0,1]$ is uncountable, there exists an element $a \in (0,1)$ such that $C \cap [a,1]$ is uncountable.
Indeed, namely any element $a$ such that $$a < \operatorname{sup}C.$$ In such a case, $$C \cap [a,1] = [a,\operatorname{sup}C) \quad \textrm{or} \quad C \cap [a,1] = [a,\operatorname{sup}C],$$ depending on whether $$\operatorname{sup}C \in C.$$ I have shown already that any open interval $(x_1,x_2)$ in $\mathbb{R}$ is uncountable. So, $$(a, \operatorname{sup}C) \subset [a, \operatorname{sup}C) \subset [a, \operatorname{sup}C],$$ and $[a,\operatorname{sup}C)$ and $[a,\operatorname{sup}C]$ are uncountable.
Moreover, considering the set $A$ of all elements $a \in (0,1)$ such that $C \cap [a,1]$ is uncountable, and $$\alpha := \operatorname{sup}A.$$
Then, $C \cap [\alpha,1]$ is uncountable $iff$ $$\alpha < \operatorname{sup}C.$$ If $\alpha = \operatorname{sup}C,$ then the intersection will contain one element or no elements, depending on whether $$\operatorname{sup}C \in C.$$
Is all this true?
Your proof is wrong. A way to prove the claim is as follows:
Assume not; for each $a \in (0,1)$, $C \cap [a, 1]$ is not uncountable. This implies that for each $n \in \mathbb{N}$, $A_n = C \cap \left(\dfrac{1}{n+1}, \dfrac{1}{n}\right]$ is countable. Additionally, let $A_\infty \equiv C \cap \{0\}$, which is obviously countable. But then, (including $A_\infty$ in the union)
$$C = \bigcup_{n=1}^{\infty} A_n$$
is a countable union of countable sets, an easy injection can be made from $C \to \mathbb{N^2} \to \mathbb{N}$.