We call the number $x$ an $n$-digit number if it can be written as $x = \sum_{i=0}^{n} a_i s^i$ where $\forall 0 \le i \le n$, $a_i$ is either $0$ or $1-s$ (note that $a_n$ can be $0$ too). Using the operations presented in the problem, starting from $a$, we can obtain numbers of the form $a s^N + b$ where $b$ can be any $(N-1)$-digit number. For any $c \le 1$ and $\epsilon \in \mathbb{R}^+$ we are looking for $b, N$ such that:
$$|(a s^N + b) - c| < \epsilon$$
To prove this, it is sufficient to find $b$ such that $|b-c|<\epsilon$ and then choose sufficiently large $N$ to satisfy the inequality.
We will prove by induction that for every $c \le 1$ there exists an $n$-digit number like $b$ such that
$$|b-c| < s^{n-1}$$
This obviously proves the main statement.
For the base of the induction, $n=1$ is obvious. Now let's assume the statement holds for $n$. We will prove it also holds for $n+1$. If $c > 1-s$ we define $c' = \frac{c-(1-s)}{s}$. This number is in the interval $[0,1]$ because:
$$\frac{c - (1 - s)}{s} \le 1 \iff c \le 1 - s + s = 1$$
By the induction hypothesis there exists an $n$-digit number $b$ such that:
$$\left| \frac{c - (1 - s)}{s} - b \right| \le s^{n-1} \implies \left| c - \left( (1 - s) + s \times b \right) \right| \le s^n$$
However $1 - s + s \times b$ is an $(n+1)$-digit number. If $c \le 1 - s$ then we define $c' = \frac{c}{s}$ this number is also in the interval $[0, 1]$ because:
$$c \le 1 - s \le s \implies c' \le 1$$
Then by the induction hypothesis there exists an $n$-digit number $b$ such that:
$$|\frac{c}{s} - b| \le s^{n-1} \implies |c - s \times b| \le s^n$$
Here $s \times b$ is an $(n+1)$ digit number. Thus the inductive step is proven. The proof is now complete.

$$f(x) = s x, \quad g(x) = s x + 1 - s.$$
We know that the images of functions $f, g$ on $[0, 1]$ are $[0, s]$ and $[1 - s, 1]$, respectively. It is clear that the length of each interval is $s$ and the union of them is $[0, 1]$.
At each time, the position of the insect can be obtained from the compositions of $f, g$. Hence, the possible positions of her after $n$ steps would be of the form $h_1(h_2(\dots(h_n(x)\dots)))$, where $h_i \in \{f, g\}$. It turns out that the length of the interval of each such function is $s^n$.
We then prove by induction that for each $n$ the union of the images of these intervals covers $[0, 1]$. The base is true. Assume that the statement holds true for all integers don't exceed $k$. Hence, the images of all possible compositions $h_1(h_2(\dots(h_k(x)\dots)))$ covers $[0, 1]$. Notice that image of any of $h_1(h_2(\dots(h_k(g(x))\dots)))$ or $h_1(h_2(\dots(h_k(f(x))\dots)))$ covers the image of $h_1(h_2(\dots(h_k(x)\dots)))$. this completes our proof.
Now, suppose a point $a \in [0, 1]$. We shall then prove that there is a positive integer $n$ that the image of $h_1(h_2(\dots(h_n(x)\dots)))$ contains $a$. For this reason, choosing $n$ such that $\frac{1}{2022} > s^n$ it follows that the insect shall be in the interval of $(a - \frac{1}{2022}, a + \frac{1}{1402})$. We are done. ■