For a given positive integer $n$ let $S_n$ denote the set of all possible values that $x_n$ can obtain for different choices of numbers $a_i$, $1 \le i \le n$. For example:
$$S_1 = \{\frac{1}{2}, 2\}, \quad S_2 = \{\frac{1}{3}, \frac{2}{3}, \frac{3}{2}, 3\}, \quad S_3 = \{\frac{1}{4}, \frac{2}{5}, \frac{3}{5}, \frac{3}{4}, \frac{4}{3}, \frac{5}{3}, \frac{5}{2}, 4\}, \dots$$
Note that, for every $x \in S_n$ both $x+1$ and $\frac{1}{x+1}$ belong to $S_{n+1}$. Furthermore, one of those two numbers is smaller than 1 and the other is greater than 1. Thus, each $S_n$ consists of even number of numbers. For a given positive integer $n$ let
$$S_n = \{a_1, a_2, \dots, a_{2m}\}, \text{ with } a_1 < a_2 < \dots < a_{2m}$$
and
$$S_{n+1} = \{b_1, b_2, \dots, b_{2k}\}, \text{ with } b_1 < b_2 < \dots < b_{2k}.$$
The following statements can easily be proved by induction:
* Claim 1. $m = 2^{n-1}$, $k = 2^n$ and thus $|S_{n+1}| = 2|S_n|$.
* Claim 2. $a_1 = \frac{1}{n+1}$, $a_m = \frac{n}{n+1}$, $a_{m+1} = \frac{n+1}{n}$, $a_{2m} = n+1$.
* Claim 3. $b_1 < b_2 < \dots < b_{2m} < 1 < b_{2m+1} < \dots < b_{4m}$.
* Claim 4. $a_i = \frac{1}{a_{2m+1-i}}$, $b_i = \frac{1}{b_{4m+1-i}}$.
* Claim 5. $b_i = \frac{1}{1 + a_{2m+1-i}}$, for $1 \le i \le 2m$.
* Claim 6. $b_i = \frac{1}{b_{2m+1-i}}$, for $1 \le i \le 2m$.
Combining Claim 5 and Claim 4 we get:
$$b_i = \frac{a_i}{1 + a_i}, \quad \text{for } 1 \le i \le 2m. \qquad (2)$$
For all integers $n \ge 2$, we will prove that $a_{i+1} - a_i \le \frac{1}{2n-1}$ for all $1 \le i \le m$. That is obviously true for $n=2$.
As an induction hypothesis, let us assume that this holds for $n$. We will prove that the claim holds for $n+1$, i.e. we will prove that $b_{i+1} - b_i \le \frac{1}{2n+1}$ for all $1 \le i \le 2m$.
From (2) and Claim 2 we know that
$$b_m = \frac{n}{2n+1}, \quad b_{m+1} = \frac{n+1}{2n+1}.$$
Therefore,
$$b_{m+1} - b_m = \frac{1}{2n+1},$$
thus the claim holds for $i = m$. Let us prove that the claim holds for $i < m$. We have:
$$b_{i+1} - b_i = \frac{1}{a_{i+1} + 1} - \frac{1}{a_i + 1} = \frac{a_{i+1} - a_i}{(1+a_i)(1+a_{i+1})}.$$
By induction hypothesis, $a_{i+1} - a_i \le \frac{1}{2n-1}$ and from Claim 2 we know that $a_i \ge \frac{1}{n+1}$ and $a_{i+1} \ge \frac{1}{n+1}$. Thus,
$$b_{i+1} - b_i = \frac{a_{i+1} - a_i}{(1+a_i)(1+a_{i+1})} \le \frac{\frac{1}{2n-1}}{(1+\frac{1}{n+1})(1+\frac{1}{n+1})} < \frac{1}{2n+1},$$
with the last inequality being equivalent to $2n^2 - 5 > 0$ which holds for $n \ge 2$. Finally, we will prove that $b_{i+1} - b_i \le \frac{1}{2n+1}$ for $m < i < 2m$.
From Claim 6 we have:
$$b_{i+1} - b_i = \frac{1}{b_j} - \frac{1}{b_{j+1}} = \frac{b_{j+1} - b_j}{(1+b_j)(1+b_{j+1})} < b_{j+1} - b_j < \frac{1}{2n+1}.$$
where we denoted $j = 2m - i < m$.
For $n = 101$, let $S_{101} = \{c_1, c_2, \dots, c_s\}$, where $c_1 < c_2 < \dots < c_s$.
Since $c_1 = \frac{1}{102}$ and $c_s = \frac{101}{102}$ we know that $c_1 - \frac{1}{111} < \frac{1}{201}$ and $\frac{110}{111} - c_s < \frac{1}{201}$.
Therefore, in set $S'_n = \{c_0, c_1, \dots, c_s, c_{s+1}\}$ with $c_0 = \frac{1}{111}$ and $c_{s+1} = \frac{110}{111}$ the following holds:
$$c_{i+1} - c_i \le \frac{1}{201}, \text{ for } 0 < i \le s.$$
For given $x \in [\frac{1}{111}, \frac{110}{111}]$, let $j$ be an index such that $c_j \le x \le c_{j+1}$. Since, $c_{j+1} - c_j \le \frac{1}{201}$ we conclude that $|x - c_j| \le \frac{1}{402}$ or $|x - c_{j+1}| \le \frac{1}{402}$.