> 
> Question: Let $I\_1,I\_2,I\_3$ be three open intervals of $\mathbb{R}$ such that none is contained in another. If $I\_1\cap I\_2\cap I\_3$ is non-empty, then show that at least one of these intervals is contained in the union of the other two. 
> 
> 
> 

My approach: Let us assume the contrary for the sake of contradiction, that is we have $$I\_1\nsubseteq I\_2\cup I\_3, \\I\_2\nsubseteq I\_3\cup I\_1\text{ and} \\I\_3\nsubseteq I\_1\cup I\_2.$$

Now, I can observe that if we assume so, then $\exists i,j$ such that $i\neq j$ and $I\_i\subseteq I\_j$, which is a contradiction to the statement which states that none of the intervals $I\_i$ is contained in another interval $I\_j$. 

But how to formally prove the same?