First prove for $6-gon$ and $8-gon$. Then use the fact that $100=6\cdot 2+8\cdot 11.$ Will add details later..

@above I can add details for you. First we will show that we can pick 4 points inside of any convex $8$-gon, so that each vertex will be on a straight line connecting some two marked points. Proof: Name the vertexes from $1$ to $8$ in clockwise order and mark the intersections $((1,6),(3,8)),((1,6),(4,7)),((2,5),(3,8)),((2,5),(4,7))$. Now we will show that between 7 points we can pick 6 of them such that we can mark 3 points, so that selected 6 points will be on a straight line connecting some two marked points. Proof: Name the vertexes from $1$ to $6$ in clockwise order and mark the intersections $((1,4),(2,5)),((1,4),(3,6)),((2,5),(3,6))$ if they are not concurrent. Otherwise mark the intersections $((1,4),(2,5)),((1,4),(3,7)),((2,5),(3,7))$ since we know that these three are not concurrent. Now we finish by marking $3$ points for $6$ points on the $100$-gon $2$ times, then marking $4$ points for $8$ points on the $100$-gon $11$ times and we are done.