Using union of countably infinite sets, I tried to prove that set of all real numbers in [0,1) is countable Cantor's diagonal method shows that the set $S=\{x\in \Bbb R|x \in [0,1)\}$ is uncountably infinite, because there is no bijection between the set $S$ and the set of natural number $\Bbb N$.
I came up with this method of mapping the set $S$ to $\Bbb N$. It should be wrong, but I don't know where.
We start with any $x \in S$, and we will build the whole set $S$ using the following method. First, for example, choose $x$ that $$x=0.12345678900000......$$ Next, we can choose any digit of x, and change it such that $0 \rightarrow 1$, $1 \rightarrow 2$, $2 \rightarrow 3$, $...$, $8 \rightarrow 9$, $9 \rightarrow 0$, to make a new real number $x_1\in S$. Because the list of digits of $x$ is countably infinite, there are countably infinite ways of choosing a digit, making the set $\chi_1$ of all real numbers differs $x$ by 1 digit a countably infinite set.
For any $x_1^i \in \chi_1$ (the subscript "1" stands for "differs $x$ by 1 digit"), we can follow the previous method, to make a set of real numbers differing $x$ by 2 digits (actually $x$ itself is in this set too, but we don't care about duplication). There are countably infinite number of such sets, denoted $\chi_2^i$, built from each number $x_1^i$ of the set $\chi_1$.
At this point, we might attempt to create the union of all the sets that we have just created: $$U_2=\chi_1 \cup \biggl( \bigcup_{i=1,\\i\in \Bbb N}^{\infty}\chi_2^i\biggr)$$ $U_2$ is countably infinite.
We can now create the set $U_3$ as the countably infinite union of all countably infinite sets $\chi_3^i$. Then $U_4$, $U_5$, ...
For any number $r$, $r \in S$, it differs $x$ atmost at countably infinite number of digits. Therefore, by repeating this procedure countably infinitely many times, we can construct a set that is a superset of $S$ $$U_\infty = U_{i\in \Bbb N,i\rightarrow \infty} \supset S$$ $U_\infty$ is a proper superset of S, because it  contains infinitely many duplicates of any number $r \in S$. But if $U_\infty$ is countably infinite, then S is also countably infinite.

Please help me see where the error in my "proof" is.

 A: Your claim that $S\subset \bigcup_{n\in\mathbb N} U_n:=U_\infty$ is false. I believe this may stem from a misunderstanding of what an infinite union is. In elementary set theory we have, for any index set $I$ and collection $\{A_i\}_I$, that $y\in \bigcup_{i\in I}A_I$ if and only if there exists some $\alpha\in I$ such that $y\in A_\alpha$.
To prove that your claim is false, let $y\in S$ be some real that differs from $x$ at every digit. Suppose that $y\in U_\infty$. By definition we must then have that $y\in U_n$ for some $n\in \mathbb N$, but then $y$ can only differ from $x$ at at most $n$ digits, which is a contradiction.
This is akin to a misunderstanding often seen in mathematical induction. For example with induction you can prove, for every $n\in \mathbb N$, that $\int_0^n 1dx<\infty$. However you cannot conclude from this that $\int_0^\infty 1dx<\infty$. In other words induction does not allow you to pass to the limit. In the same way an infinite union does not allow you to "pass to the limit" to form some set $U_{\mathbb N}$ containing numbers differing from $x$ at all decimal points. It only allows you to talk about numbers differing from $x$ at any arbitrary but finite number of decimal points. If you want to pass to the limit, then you are no longer just taking the countable union of countable sets anymore.
