TITLE: Always a planar-drawn cycle through $n$ points
QUESTION [8 upvotes]: Given $n$ points in the plane, can we always find a cycle through all of them that has only straight line edges and no edges intersect (planar-drawn)?

Intuitively the answer is yes, but I am struggling with a proof. In the example above, the $2^{nd}$ graph demonstrates an incorrect algorithm, whereas the last graph demonstrates a working algorithm. I have tried using induction.
For different variants there are constraints - this is not always possible if the points are colored (e.g. if two points of one color lies on non-adjacent points of the convex hull and any path between them partitions the other color).

REPLY [2 votes]: The answer is yes.
First we draw a cycle with the outmost vertices such that other vertices are within the cycle. (This is step 2 in above picture.)
Then we repeat this for vertices that are not on the first cycle and draw the second cycle.
We repeat this until all vertices lie on a unique simple closed cycle.
Now we delete an edge from each two adjacent cycles and connecting vertices such that two cycles exchange in one longer cycle. Then repeat this until we have had a cycle consists all vertices.