Prove that on space of segments $I=[a,b]$ $\rho$ is a metric I want to prove that we can equip space of segments $I=[a,b]$ with a metric
$\rho (I_1,I_2)=|I_1|+|I_2|-2|I_1 \cap I_2|$, where $|I|$ is length of $I$.
It's easy to prove that $\rho (I_1,I_2) >0$, because $|I_1 \cap I_2| < |I_1|,|I_2|$ (we do not consider the obvious situation when $I_1=I_2$). Therefore, $$\rho (I_1,I_2)=|I_1|+|I_2|-2|I_1 \cap I_2|=|I_1|+|I_2|-|I_1 \cap I_2|-|I_1 \cap I_2|>|I_1|-|I_1|+|I_2|-|I_1 \cap I_2|>|I_1|-|I_1|+|I_2|-|I_2|=0$$
My glitch is on triangle inequality. But I have an idea...
The triangle inequality is the following (we need to prove it):
$$\rho(I_1,I_2)+\rho(I_2,I_3) \ge \rho(I_1,I_3)$$
$$|I_1|+|I_2|-2|I_1 \cap I_2| + |I_2|+|I_3|-2|I_2 \cap I_3|
 \ge |I_1|+|I_3|-2|I_1 \cap I_3|$$
$$|I_2|+|I_1 \cap I_3|\ge |I_1 \cap I_2|+|I_2 \cap I_3|$$
 Hmm, maybe the idea is more tricky, e.g. we can consider case when $|I_1 \cap I_2|<\frac{|I_1|}{2}$ and $|I_2 \cap I_3|<\frac{|I_3|}{2}$.  
 A: Your metric may be written:
$$ \rho(I_1,I_2)= |I_2\setminus I_1| + | I_1 \setminus I_2|$$
which facilitates calculations. For example, since $A\cap B^c \subset (A\cap C^c) \cup (C\cap B^c)$ we have
$$|I_3\setminus I_1| \leq |I_3\setminus I_2|+|I_2\setminus I_1|$$ which immediately yields the triangle inequality.
Another more analytic approach  is to use characteristic functions on intervals and
write $$\rho(I_1,I_2)= \|1_{I_1} - 1_{I_2}\|_{L^1}$$ which also gives the triangular inequality straightaway. 
A: You’re trying to prove that
$$|I_1|+2|I_2|+|I_3|-2|I_1\cap I_2|-2|I_2\cap I_3|\ge|I_1|+|I_3|-2|I_1\cap I_3|\;,$$
which can be simplified to
$$|I_2|-|I_1\cap I_2|-|I_2\cap I_3|\ge-|I_1\cap I_3|$$
and thence to
$$|I_2|+|I_1\cap I_3|\ge|I_1\cap I_2|+|I_2\cap I_3|\;.$$
Let $J_1=I_1\setminus(I_1\cap I_3)$ and $J_3=I_3\setminus(I_1\cap I_3)$; each of these sets is either an interval or a union of two disjoint intervals, so $|J_1|$ and $|J_3|$ are meaningful. Clearly
$$|I_1\cap I_2|+|I_2\cap I_3|=\color{brown}{|J_1\cap I_2|+|(I_1\cap I_3)\cap I_2|+|J_3\cap I_2|}+\color{blue}{|(I_1\cap I_3)\cap I_2|}\;.$$
Now 
$$\color{blue}{|I_1\cap I_2\cap I_3|}\le|I_1\cap I_3|\;,$$
and
$$\color{brown}{|J_1\cap I_2|+|(I_1\cap I_3)\cap I_2|+|J_3\cap I_2|}\le|I_2|\;,$$
since the sets $J_1\cap I_2,I_1\cap I_2\cap I_3,$ and $J_3\cap I_2$ are pairwise disjoint subsets of $I_2$, so
$$|I_1\cap I_2|+|I_2\cap I_3|\le|I_2|+|I_1\cap I_3|\;,$$
as desired.
