Show that at least one of the intervals is contained in the union of the other two. 
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?
 A: Let $I_i=(a_i,b_i)$ and let $x\in I_1 \cap I_2 \cap I_3$. Let also $a_i:=\min \{ a_1,a_2,a_3\}$ and $b_j:=\max\{b_1,b_2,b_3\}$. Since $x\in I_i\cap I_j$ we have $a_j < b_i$, in particular there is no gap between $I_i$ and $I_j$. Moreover for every $k$ we have $a_k \ge a_i$ and $b_k \le b_j$, hence $I_k\subseteq I_i\cup I_j$.
A: Let $I_i=(a_i,b_i), i=1,2,3.$ Now since $I_1\cap I_2\cap I_3\neq \phi\implies \exists x\in\mathbb{R}$, such that $x\in I_1\cap I_2\cap I_3$. 
Again since $x\in I_1\cap I_2\cap I_3$, implies that $x\in I_i\cap I_j, \forall i\neq j.$
Now let $a'=\min\{a_1,a_2,a_3\}$ and let that $b'=\{b_1,b_2,b_3\}$. 
Now since none of $I_i$ is contained in $I_j$, where $i\neq j$, implies that, we have $$(a_i,b_i)\nsubseteq (a_j,b_j), \forall i\neq j.$$ 
We also have that $x\in I_i\cap I_j\implies I_i\cap I_j\neq \phi.$ Therefore, either we have $a_i<a_j<b_i<b_j$ or we have $a_j<a_i<b_j<b_i$. 
This implies that $\forall i,j,$ we have $a_i<b_j$. 
The conclusion $a_i<a_j<b_i<b_j$ or $a_j<a_i<b_j<b_i, \forall i\neq j$, helps us in concluding that if $a'=a_k$ and $b'=b_l$, then we must have $k\neq l$. 
This implies that $a_k<a_l<b_k<b_l$. Now let the other set be $(a_m,b_m)$. 
Therefore, we have $a_k<a_m<b_k<b_m\implies a_k<a_m<b_k<b_m<b_l$. 
Now two cases are possible, either we will have $a_k<a_m<a_l<b_k<b_m<b_l$ or we will have $a_k<a_l<a_m<b_k<b_m<b_l$. 
Observe that if any of the above mentioned cases is true then we can conclude that $(a_m,b_m)\subseteq(a_k,b_k)\cup (a_l,b_l)$. 
This implies that at least one of the given sets is contained in the union of the other two.  
A: Suppose without loss og generality that $I_1$ is the interval with the left endpoint most to the left and $I_3$ is the interval with the right endpoint most to the right. Since $I_1 \cap I_3$ is non-empty, $I_1 \cup I_3$ is an interval that has to contain $I_2$.
