So $x = 0.a_1a_2a_3...a_na_{n+1}a_{n+2}...$
$a_n = 1$ if $n = 10^k$ $k\in\mathbb{N} \cup\{0\}$
$a_n = 0$ if $n \neq 10^k$ for any $k\in\mathbb{N} \cup\{0\}$
It's obvious that $x$ is not rational since it doesn't have period, because for any $n\in\mathbb{N}$
$10^n - 10^{n-1} \neq 10^{n-1} - 10^{n-2}$
This means that number of zeros between "1"s is always different.
$x$ would be rational though if
$\forall l\in\mathbb{N}\cup\{0\}$
$\exists N\in\mathbb{N}$
$\exists m\in\mathbb{N}$ such that
$$\{a_N,a_{N+1},\cdots ,a_{N+m}\}=\{a_{N+lm},a_{N+lm+1},\cdots a_{N+lm+m}\}$$
So I described what this sequence looks like, but I have no clue, what to do next. My intuition says that irrationality of $x$ should be proved using "contradiction method", which implies assumption, that $x$ is rational, hence it has period, therefore some sequence of $a$s is not a subset of an another larger sequence of $a$s with different indexes. But how to prove it and how to write it mathematically is unfortunately unclear.