Is this a paradox in the Cantor set? In a previous question here Can we define the Cantor Set in this way?
we defined a family of sets $ \left\{ C_0,C_1,C_2,C_3,\dots \right\}$
We can call this set $S_1$ , where the values of these elements is
$C_0 = \left\{ 0.0   \right\}$
$C_1 = \left\{ 0.0 , 0.2  \right\}$
$C_2 = \left\{ 0.0 , 0.2 ,0.02 , 0.22 \right\}$
$C_3 = \left\{ 0.0 , 0.2 ,0.02 , 0.22 ,0.002 , 0.202 ,0.022 , 0.222 \right\}$
and so on...
We can then take the union of $\bigcup S_1$ and get some countable set $X$
We will define a set  $S_2$ to be all the members of  $S_1$ that are not redundant.
$C_0$ is redundant because it can be removed from $S_1$ and the union will still be the same.
In fact all the elements of $S_1$ are redundant so our set
$\bigcup S_2 = \left\{ \not C_0,\not C_1,\not C_2,\not C_3,\dots   \right\}  = \varnothing \neq X \to$ Contradiction
We removed all the sets from the union that did not contribute any information, and all the information disappeared.
What went wrong here?
 A: First, note that this isn't really a question about the Cantor set, just about the combinatorics of infinite sets in general.
E.g. we can recast it in terms of sets of natural numbers: let $A_n=\{1, 2, 3, ..., n\}$, and think about $\bigcup A_n$. Or, even easier, we could just take $B_n=\{0\}$ (that's not a typo) and think about $\bigcup B_n$.

Now as to the answer to your question, nothing went wrong except your intuition for how infinite sets behave. Even though removing any finite number of the $A_n$s leaves the union of the rest unchanged, that's no reason to believe that removing infinitely many of the $A_n$s should do the same. There are lots of times we see a "phase transition" between finite behavior and infinite behavior - e.g. changing finitely many terms of a divergent series to $0$ leaves the result divergent, but if you change infinitely many terms to $0$ it might converge.
A: Casting this in terms of the set of natural numbers, there seems to be no contradiction if $ \mathbb{N} \in S_1 $.
Let:$$  S_1 =  \left\{ C_m =  \left\{ n \in \mathbb{N} : n \le m  \right\} : m \in \mathbb{N} \right\} $$ 
Then:$$C_0=  \left\{ 0 \right\}$$
$$C_1=  \left\{ 0,1 \right\}$$
$$C_2=  \left\{ 0,1,2 \right\}$$
$$\dots$$
$$ S_1 = \left\{ C_0,C_1,C_2,C_3,\dots , \mathbb{N} \right\} $$
$$ \mathbb{N} \in S_1 $$
$$\bigcup S_1 = \mathbb{N}  $$
Let:$$ S_2 = \left\{ C_n\not\subset (X \in  S_1) : n \in \mathbb{N}  \right\} $$
Then:$$ S_2 = \left\{ \not C_0,\not C_1,\not C_2,\not C_3,\dots , \mathbb{N} \right\} $$
$$S_2 = \left\{\mathbb{N} \right\} $$
$$\bigcup S_2 = \mathbb{N}  $$
It only gets ugly if $ \mathbb{N} \notin S_1 $
