Taiwan IMO Selection Camp

Suppose there exist $0 < a < b < 1$ such that there is no positive integer $n$ making $(a_n - a_{n+1})(b_n - b_{n+1}) < 0$.
$$f(x) - x \begin{cases} > 0 & \text{if } x < \frac{1}{2}, \\ < 0 & \text{if } x \ge \frac{1}{2}. \end{cases}$$
Then for a given positive integer $n$, it must be that $a_n, b_n \in (0, \frac{1}{2})$ or $a_n, b_n \in [\frac{1}{2}, 1)$.
Let $d_n = b_n - a_n$, then if $a_n, b_n \in (0, \frac{1}{2})$ we have $d_{n+1} = d_n$, otherwise we have
$$d_{n+1} = d_n(a_n + b_n) \ge d_n(1 + d_n),$$
so $d_n$ is an increasing sequence.
Suppose $0 < a_n < b_n < \frac{1}{2}$, then $\frac{1}{2} < a_n < b_n < 1$. In any case we have $d_{n+2} \ge d_n(1+d_n)$.
Hence $d_{2m+1} \ge d_1(1+d_1)^m \ge d_1(1+md_1)$.
There exists a sufficiently large $m$ such that
$$d_1(1+md_1) > 1,$$
but this is impossible. Hence there exists a positive integer $n$ such that $(a_n - a_{n+1})(b_n - b_{n+1}) < 0$.