A convex closed plane curve never intersects itself? Intuitively, I think that a convex closed curve has to be simple (i.e. cannot intersect itself except at the starting and the ending points). How one can prove it rigorously?
My attempt: Suppose it is a convex closed curve and it intersects itself. There exists(?) a neighborhood that the curve is not convex anymore, leading to contradiction.   
Also, do we need any smoothness condition?
 A: The problem here is that there are two notions of "curve"; one is a subset of the plane homeomorphic to the circle; the other is the image of the unit circle under some continuous map to the plane. 
If you take the first notion, then "intersect itself" doesn't really make sense: every set intersects itself, with $X \cap X = X$. When you want to say "I mean at two different points", I say "They're the same point of the plane!"
So we have to be talking about the second case. And the problem there is that there are trivial counterexamples. For instance, the map 
$$
f: S^1 \to \Bbb R^2 : (x, y) -> x
$$
has as its image the interval from $-1$ to $1$ on the $x$-axis, which is indeed convex, but we also have $f(0, 1) = f(0, -1)$, so it clearly "intersects itself". 
Now you might say "But I meant for the image to be homeomorphic to the unit circle!", and in that case, the map
$$
g: S^1 \to \Bbb R^2 : (2xy, x^2 - y^2)
$$
which wraps the unit circle twice around itself is a counterexample. 
In short, the answer to the question you asked is "no", but the correct rephrasing of the question you probably wanted to ask is really difficult. 
