If $P$ denotes the Cantor set, then show that $[0, 1] \setminus P$ is dense in $[0, 1]$ I have done the following. Suggest if the Proof is rigorous enough.
Let us select any point $x\in [0,1]$. Ternary expansion of $x$ can be represented by $x=0.b_1 b_2 b_3 b_4\ldots$ where $b_i=0,1\text{ or }2$.
So, any $x\in [0,1]$ either lies in Cantor Set $P$ or does not lie in $P$.
Case 1: $x\notin P$
then $x\in \{[0,1]\setminus P\}\subset [0,1]$
So, $\forall r>0, B_{r}(x)$ contains an element of $[0,1]\setminus P$ since $B_{r}(x)$ contains at least $x$.
Case 2:  $x\in P$
Then $x$ has a ternary expansion $x=0.b_{1}b_{2}b_{3}b_{4}\ldots$ where $b_{i}=0\text{ or }2$.
Appending a $1$ far enough down the expansion of $x$, we get $x^{'}\in \{[0,1]\setminus P\}\subset [0,1]$. Also, $x^{'}$ is arbitrarily close to $x$.  So, $\forall r>0, B_{r}(x)$ contains $x^{'}$ which is an element of $[0,1]\setminus P$. 
So, $ [0, 1] \setminus P$  is dense in $[0, 1]$.
Is Case 2 rigorous enough?
 A: *

*Cantor sets in ternary use only 0 or 2 (instead of 1)

*you cannot "append" a 1 further enough, since the ternary expansion may require infinitely many "2" (e.g. 0.2222222$\cdots$).
To prove $[0,1]\backslash P$ is dense in $[0,1]$, just note that $P$ has empty interior since its Lebesgue measure is 0.
A: It isn’t really necessary to consider the case $x\notin P$, since any open nbhd of such a point obviously meets $[0,1]\setminus P$, but there’s no real harm in it. As noted in Milly’s answer, there are a couple of problems with your argument in the essential case in which $x\in P$. That implies that $x$ has a ternary expansion $0.b_1b_2b_3\ldots$ in which each $b_k$ is either $0$ or $2$. Changing some $b_k$ to a $1$ doesn’t quite guarantee that the new number isn’t in $P$: if $b_i=2$ for all $i>k$, then you get
$$0.b_1\ldots b_{k-1}1222\ldots=0.b_1\ldots b_{k-1}2000\ldots\;,$$
which is in $P$. However, you can change a whole tail of digits to $1$s:
$$0.b_1\ldots b_{k-1}111\ldots$$
cannot be in $P$. 
It only remains to show that if you take $k$ large enough, you can make 
$$|0.b_1\ldots b_{k-1}b_kb_{k+1}\ldots-0.b_1\ldots b_{k-1}111\ldots|$$
arbitrarily small. You might note that $x$ lies between $0.b_1\ldots b_{k-1}$ and $0.b_1\ldots b_{k-1}222\ldots$ inclusive; this allows you to express the maximum possible distance between $x$ and the new number in terms of $k$.
