Irish Mathematical Olympiad

First of all, note that $a_n = \frac{1}{n}(1 + x + \cdots + x^{n-1})$, hence
$$\begin{align*} a_{n+1} - a_n &= \frac{n(1 + x + \cdots + x^n) - (n+1)(1 + x + \cdots + x^{n-1})}{n(n+1)} \\ &= \frac{1}{n(n+1)} \left( n x^n - \sum_{k=0}^{n-1} x^k \right) \\ &= \frac{1}{n(n+1)} \sum_{k=0}^{n-1} (x^n - x^k). \end{align*}$$

Since $0 \le x < 1$, for $i \ge 0$ we have $0 \le x^i \le 1$, hence $0 \le |x^n - x^k| \le 1$ for $k = 0, 1, \dots, n-1$, and so
$$$$
$$\begin{aligned} |a_{n+1} - a_n| &= \frac{1}{n(n+1)} \left| \sum_{k=0}^{n-1} (x^n - x^k) \right| \\ &\le \frac{1}{n(n+1)} \sum_{k=0}^{n-1} |x^n - x^k| \le \frac{n}{n(n+1)} = \frac{1}{n+1}. \end{aligned}$$