Problem 1
Base case: A triangle. Trivial.
Induction: Let P (drawn in black) be any non-parallelogram polygon of n > 3 vertices, and assume the conjecture is true
for all m < n. Because P is not a parallelogram, we can select two non-adjacent, non-parallel sides of P. Draw
extensions of these two edges so that the extensions intersect at their endpoints. (Drawn in green.)
Now we delete the edges between (i.e., enclosed by) the two green edges. Let's call this the green polygon.
The green polygon has fewer than n edges, so it can be enclosed in a triangle colinear with three of its sides. Since every edge of the green polygon is colinear with an edge of P, and the green polygon encloses P. Therefore any triangle which
encloses the green polygon also encloses P.
QED
Comments
You can't just use a convex polygon which you draw, and then "add a point." This is not proving the conjecture; it only proves it for the polygon to which you added a point. Besides this, how do you "add a point"? If you want to show this for all convex polygons, you need to prove that all convex polygons are constructible from your method of adding points to polygons which were create by adding points to polygons . . . etc. You'll get an infinite recursion, and you'll need an induction inside your induction. It's not a good way to prove things.
The proper method of proof is: Suppose I have a convex polygon. Then I remove an edge, perhaps by ear-cutting. Then I do whatever I need to do with the smaller polygon (for which the conjecture holds), and then I put the edge back in!
It's subtle, but crucial. A lot of induction proofs are ruined this way, and it is possible to prove incorrect results if the logic of "adding a point" is used.
Problem 2
Another solution for this problem:
Problem 3
The conjecture is false.
No matter where s and t are, if the guards start in one of the four leaves, neither one can get out, since each leaf is only visible to the one across from it. It is a street, since you can partition it with s and t in opposite leaves, making the two chains mutually weakly visible.
Note that even if the guards start at one of the four center vertices, they still can't get out of a single leaf -- although they do have the ability to choose in which leaf they become stuck, depending on whether the right guard or the left guard moves first.
Comments
Note that you can't just pick s and t and tell me that it's not a walkable polygon. The polygon is walkable if it CAN be partitioned into two chains -- that is, if THERE EXIST s and t such that two guards can walk the polygon.
Don't confuse THERE EXISTS and FOR ALL.