Take any point $P$ on the plane, and let $\ell_\theta$ be the directed line through $P$ making an angle $\theta$ with the positive $x$-axis. Note that for the $i$-th collection ($i = 1,2$), the projection of each triangle in the $i$-th collection onto $\ell_\theta$ is a finite interval, and by the intersecting property, these intervals pairwise intersect. This means that the intersection of the projections of all triangles in the $i$-th collection is also a finite interval (this can be proven using convexity or induction¹); let $d_i(\theta)$ denote the signed distance from $P$ to the midpoint of the interval formed by the projection of the $i$-th collection (according to the direction of $\ell_\theta$).

Now, note that $d_i(\theta)$ is a continuous function of $\theta$, and $d_i(\theta + \pi) = -d_i(\theta)$. This means that if $d_1(\theta) > d_2(\theta)$, then $d_1(\theta + \pi) < d_2(\theta + \pi)$, so by the Intermediate Value Theorem there exists $\varphi \in (\theta, \theta + \pi)$ such that $d_1(\varphi) = d_2(\varphi)$, and this means that the intersection of the projections onto $\ell_\varphi$ of all triangles in both collections is nonempty. Choose any point $Q$ in this intersection; then it is easy to see that the line through $Q$ perpendicular to $\ell_\varphi$ will pass through the interiors of all triangles in both collections. The same discussion applies to the case $d_1(\theta) < d_2(\theta)$, and the case $d_1(\theta) = d_2(\theta)$ is obvious, so the proof is complete.