Consider the following algorithmic process on $A$. Find the interval with the leftmost left endpoint (break ties by choosing amongst the winners the one with the rightmost right endpoint). Call it $l_1$. Find the interval with the rightmost right endpoint (break ties by choosing amongst the winners the one with the leftmost left endpoint). Call it $r_1$. It is possible that $l_1=r_1$. Delete $l_1$ and $r_1$ from $A$. Continue until there are $l$'s and $r$'s up to $l_k$ and $r_k$ and one remaining interval $I\in A$. Observe that $k\geq m-1$ and $$I\subseteq \left(\bigcap_{i=1}^k l_i\right)\cup\left(\bigcap_{i=1}^k r_i\right)$$The result follows $\blacksquare$