$C$ = Cantor set. $C$ is closed, so $C^c$ is the countable union of disjoint open intervals. So let $$C^c = \bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} A_i $$
where $A_i$ are disjoint open intervals.
For each $i \in \mathbb{N},\ A_i\ $ differs from a closed set by just two points, namely the endpoints of the interval $A_i.$ For each $i \in \mathbb{N},\ $ let
$$ B_i = \{x_{i1}, x_{i2} \} = \{ \text{ the two endpoints of } A_i \}. $$
Then $\left|\bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} B_i\right| \leq 2 \left|\mathbb{N}\right| $ is countable. However, $\bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} B_i$ is surely the Cantor set, which is uncountable.
Something is incorrect in all of this, but what?
There are lots of points in the Cantor set other than the end points of the intervals $A_i$. Infinite sums of the type $\sum \frac {a_i} {3^{i}}$ ($a_i \in \{0,2\}$) are in $C$ and they are not end points unless the sum reduces to a finite sum.