Now let $mathbb{P}:={W:U^{-}subseteq Wmbox{ convex proper subset open in }A}$.
We claim that the p.o. $(mathbb{P},subseteq)$ has the Zorn property.
Clearly the union of any chain of convex subsets open in $A$,
is again a convex subset open in $A$. Suppose it{per contra} that
a union of a chain $mathcal{C}subseteqmathbb{P}$ is equal to $A$.
Then since $A$ is compact, there must be a finite sub(chain) $mathcal{C}'subseteqmathcal{C}$
for which $operatorname{max}(mathcal{C}')=bigcupmathcal{C}'=A$, which contracts
the properties of members of $mathbb{P}$. Thus $mathbb{P}$ is non-empty
and has the Zorn property. So fix $U^{+}$ open such that $U^{+}cap Ainmathbb{P}$ maximal.

textbf{Sub-claim 1:}
    $U^{+}cap Asubsetoperatorname{cl}_{A}U^{+}$ strictly.\
textbf{Proof (sub-claim 1):}
    Let $xin U^{+}cap A$ and $yin Asetminus U^{+}$.
    Consider the map (here we need the underlying field to contain the reals topologically)\
    begin{math}begin{array}[t]{lrclll}
			&f &: &mathbb{F} &to &V\
			  &&: &s &mapsto &s~x+(1-s)~y\
		end{array}end{math}\
    which is continuous and affine.
    Thus the set $S:={sin[0,1]:smapsto s~x+(1-s)~yin U^{+}cap A}
			=[0,1]cap f^{-1}U^{+}cap f^{-1}A)$
    is a convex. The set $f^{-1}A$ is closed convex and contains $0,1$,
    thus contains $[0,1]$. So $S=[0,1]cap f^{-1}U^{+}$ which is convex open
    in $[0,1]$. Since $0notin S$ and $1in S$, then $S=(a,1]$ some $ain[0,1)$.
    Clearly $f(a)=lim_{ssearrow a}f(s)$ which is a limit of vectors from $U^{+}cap A$,
    and thus lies in $operatorname{cl}U^{+}$. It also clearly lies in $A$,
    since $f^{-1}Asupseteq[0,1]$. We also have $f(a)notin U^{+}$.
    Thus $operatorname{cl}_{A}U^{+}setminus U^{+}
			=Acapoperatorname{cl}U^{+}setminus U^{+}
			supseteq{f(a)}$.
    Thus the containment is strict.
QED (sub-claim 1)

textbf{Sub-claim 2:}
    If $Wsubseteq A$ is convex, then $U^{+}cup W$ is convex.\
textbf{Proof (sub-claim 2):}
    Fix $xin U^{+}cap A,tin(0,1)$. Consider the map\
    begin{math}begin{array}[t]{lrclll}
			&T &: &V &to &V\
			  &&: &y &mapsto &t~x+(1-t)~y\
		end{array}end{math}\
    This is continuous and affine. By the proposition below,
    we also see that $T(operatorname{cl}(U^{+}))subseteq U^{+}$.
    So $Acap T^{-1}U^{+}$ is convex, open in $A$, and contains $Acapoperatorname{cl}(U^{+})$
    which strictly contains $U^{+}cap A$ by Claim 1. Thus by maximality of $U^{+}$ in $mathbb{P}$,
    $Acap T^{-1}U^{+}overset{mbox{tiny must}}{=}A$. In particular, $T^{-1}U^{+}supseteq A$,
    and so $TWsubseteq TAsubseteq U^{+}$. Utilising such maps as $T$,
    we see that a convex-linear combination of any pair of
    elements from $U^{+}cup W$ is contained in $U^{+}cup W$.
QED (sub-claim 2)

textbf{Sub-claim 3:}
    $Asetminus U^{+}$ is a singleton.\
textbf{Proof (sub-claim 3):}
    Else, let $x_{1},x_{2}in Asetminus U^{+}$ distinct.
    Let $W$ be a convex open set separating $x_{1}$ from $x_{2}$.
    Then by Claim 3, $U^{+}cap Acup Wcap A$ is convex open in $A$.
    Since $x_{1}in Wsetminus U^{+}$, this convex set is strictly large than
    $U^{+}cap A$. By maximality of $U^{+}$ in $mathbb{P}$,
    we have that $U^{+}cap Acup Wcap A=Ani x_{2}$. But this contradicts
    the fact that $x_{2}notin U^{+}cup W$.
QED (sub-claim 3)

At last, we claim that the point $ein Asetminus U^{+}$ is extreme in A.
Consider $x,yin A$
and $tin(0,1)$ for which $tx+(1-t)y=e$. Case by case, we see
that if $x,yin U^{+}$ then $e=tx+tyin U^{+}$.
If $xin Asetminus U^{+}$, $y=frac{e-tx}{1-t}=frac{e-te}{1-t}=e=x$.
If $yin Asetminus U^{+}$, then $x=e=y$ similarly.
Thus $e$ is extreme.