Extreme points of a compact set, not compact. This is a problem from chapter 3, Convexity of Rudin's "Functional Analysis".
The problem says:
Let $$ be the smallest convex set in $\mathbb{R}^3$ that contains the points $(1,0,1)$,$(1,0,−1)$ and $(cos\phi,sin\phi,0)$, for $0\leq\phi\leq 2\pi$. Show that $$ is compact but that the set of all extreme points of  is not compact. Does such an example exist in $\mathbb{R}^2$
This question is also here: Compact set and its extreme points, but the answers given did not satisfy me and the post is inactive, so I decided to ask this problem again.
I have already shown that $K$ is compact. For me it is really clear that $ext(K)$ has all the original points except for one in the circle, and hence it is clear that $K$ is not compact. But I do not know how to do this formally. It is easy to show that $(1,0,0)$ is not an extreme points but I do not know how to prove that the rest of the original points are extreme points.
I think that in $\mathbb{R}^2$ this can not happen, because, as someone says in the post Compact set and its extreme points, the open segments will be open in $\partial K$, and hence the extreme points will be closed in $\partial K$. This idea is really intuitive for me but again I do not know how to formalize it.
I would be grateful for any idea.
 A: I am not sure exactly what your question is. It's pretty routine to analyze the double cone in Rudin's exercise. If you want a proof of the fact that if $K$ is closed in $\mathbb R^2,$ the set $\hat K$ of its extreme points is closed, here is a sketch: 
Suppose $x\notin \hat K.$ 
$1).\ $If $x\notin K$ then since $K$ is closed, there is an open set containing $x$ disjoint from $K$, hence from $\hat K$.
$2).\ $If  $x\in K$, there exist $y\neq z\in K$ such that $x=(1-t)z+ty$ for some $0<t<1.$ If any point on this segment is in $\text{int}\ K,$ then $x\in \text{int}\ K$, so there is an open set $U\subseteq K$ containing $x$ and clearly $U\cap \hat K=\emptyset.$ 
$3).\ $ If the segment $(1-t)z+ty$ is contained in $\partial K$, then (here is where we use the fact that we are working in $\mathbb R^2$) we may choose a point $w\in \text{\int}\ K$ and an open triangle $T$ with vertices $w, A, B$ such that $A$ lies on the line segment that extends the segment passing through $\overline{wz}$ and $B$ lies on the line segment that extends the segment passing through $\overline{wy}.\ T\cap \hat K=\emptyset$ by construction. 
It follows that $x$ lies in an open set disjoint from $\hat K$ and so $\hat K$ is closed because its complement is open.
