How do I show that every number of form $0.a_1a_2 \dots $ where $a_n \in \{0, 2 \}$ lies in Cantor set?
From other posts, it's clear to me that $k$-th iteration, in ternary representation, removes interval with $1$ on $k$-th position i.e. $(0.a_1 \dots a_{k-1} 1 , 0.a_1 \dots a_{k-1} 1\bar 2)$. Could it be possible to show (inductively) that every sequence of $0.a_1a_2 \dots $ where $a_n \in \{0, 2 \}$ for all $n \in \mathbb N$ lies in cantor set?
That depends a bit on how you define the Cantor set. There are at least three(*) different equivalent definitions of which using ternary expansion is one of them. To see that they are equivalent we can reformulate them so they all will produce a sequence of sets and the Cantor set being the intersection of these sets - you then prove the equivalence by showing that the definitions produce identical sequence of sets.
To reformulate the definition using ternary expansion you define:
$$C_n = \left\{\sum c_j3^{-j}\right\}$$
where $c_j \in \{0,2\}$ if $j\le n$ and $c_j\in \{0,1,2\}$ if $j\gt n$. You then by induction show that $C_j$ is the same sequence of sets as for the other definitions. To complete the proof you show that $\bigcap C_j$ is the set of numbers with ternary expansion of only $0$s and $2$s.
(*) The other two are those which start with the unit interval and either subsequently removing the mid thirds or replace the set with two copies of size $1/3$ of the previous.