[Proof] Without loss of generality, let the line segment of length 1 be $[0,1]$. Let $M$ be the union of points on the marked intervals, and $N=[0,1] \backslash M$, i.e., $N$ is the union of points on some intervals in $[0,1]$ that have no common interior. Suppose $k$ is a natural number and $1 \leqslant k \leqslant 10$, and let $x_{k}$ and $y_{k}$ denote the sum of the lengths of the intervals formed by $M \cap\left[\frac{k-1}{10}, \frac{k}{10}\right]$ and $N \cap\left[\frac{k-1}{10}, \frac{k}{10}\right]$, respectively. Since any two points in $M$ are not 0.1 units apart, when $k=1,3,5,7,9$, $y_{k} \geqslant x_{k+1}$. When $k=2,4,6,8,10$, $y_{k} \geqslant x_{k-1}$. From $\sum_{k=1}^{10}\left(x_{k}+y_{k}\right)=1$, it follows that the sum of the lengths of the marked intervals $\sum_{k=1}^{10} x_{k} \leqslant \frac{1}{2}$.