Let $\mathscr{F}$ be a finite family of open or closed intervals in the line $\mathbb{R}^1$. Show by an elementary proof (without referring to Helly's theorem), that if any $2$ of the intervals intersect, then all of them intersect. (Note: any $2$ of them does not mean only $2$ is sufficient, but that any $2$ chosen from the family must intersect.)
Here is my attempt:
Since $\mathscr{F}$ is a finite family, let it have cardinality $n$, and let's list the intervals in the family as $F_1,F_2, \dots , F_n$ such that $\inf F_i \le \inf F_{i+1}$ (note that in the case when some number of infimums of some intervals are equal, then just arbitrarily assign them continuing values of the sequence). Now, from the hypothesis we can suppose that $F_1 \cap F_i \not = \emptyset$ for $i=2, \dots , n$. Then $\sup F_1 \ge \inf F_2$ in order for $F_1$ to intersect $F_2$. Since $F_1$ intersects all other intervals as well, we must have that $\sup F_1 \ge \inf F_i$, and in particular, $\sup F_1 \ge \inf F_n$. Similarly, $\sup F_i \ge \inf F_n$ for $i=2,3,\dots , n-1$.
Now, the region (point) between (shared by) $\inf F_n$ and $\min \{\sup F_i$ : $i=1,2,\dots , n-1$} is a region (point) of intersection for all $F_i$ so that $\bigcap \mathscr{F} \not = \emptyset$
Is this proof correct? Also, would it be possible to do a proof by induction?