= Solution
A point $x$ of a <convex set> $K$ is an <extreme point> if $x=(1-t)a+tb$ with $a,b\in K$ and $0<t<1$ forces $a=b=x$. It is enough to test midpoints: a nontrivial interior point of a segment is the midpoint of a smaller nontrivial segment.
Here is a proof of the <Krein-Milman theorem>. The empty-set case is immediate, so take $K\ne\varnothing$. A <face of a convex set> is a convex subset $F\subseteq K$ such that an interior convex combination lying in $F$ has both endpoints in $F$. Consider nonempty compact faces. They include $K$. Every inclusion chain has nonempty intersection, because its closed subsets of the compact set $K$ have the <finite intersection property>. The intersection remains a compact face. The <Zorn lemma>, ordered by reverse inclusion, therefore supplies a minimal nonempty compact face $F$.
If $F$ contains two distinct points, the Hausdorff locally convex hypotheses and the allowed separation theorem supply a continuous real <linear functional> $\ell$ taking different values there. Its maximizer set in $F$ is nonempty and compact. It is convex, and if a convex combination attains the maximum, both endpoint values attain it too; hence it is a face of $F$. A face of a face is a face of $K$, by applying the defining endpoint property twice. This maximizer face is proper because $\ell$ is nonconstant, contradicting minimality. Thus $F$ is a singleton, whose point is extreme in $K$. The same <minimal compact face argument> inside any nonempty compact face shows that each such face contains an <extreme point> of $K$.
Let $H=\overline{\operatorname{conv}}(\operatorname{ext}K)$. Since $K$ is compact, hence closed in the <Hausdorff space>, and convex, $H\subseteq K$. It is nonempty, closed and convex. If $p\in K\setminus H$, the allowed strict separation theorem supplies a continuous <linear functional> with
$$
\ell(p)>\sup_{h\in H}\ell(h).
$$
The maximizer face of $\ell$ on $K$ contains an <extreme point> $e$ by the preceding argument. Then $e\in H$ but $\ell(e)=\max_K\ell\ge\ell(p)>\sup_H\ell$, a contradiction. Therefore
$$
\boxed{K=\overline{\operatorname{conv}}(\operatorname{ext}K).}
$$
All closures refer to the given locally convex topology, and <compactness> is what justified both maxima and chain intersections.
For the real <l-infinity sequence space>, if $|x_j|<1$ for any coordinate, choose $0<\epsilon\le1-|x_j|$. The two distinct sequences $x\pm\epsilon e_j$ belong to the closed unit ball and have midpoint $x$, so $x$ is not extreme. Conversely, if every $x_j$ is one or minus one, writing $x=(1-t)a+tb$ for unit-ball sequences forces $a_j=b_j=x_j$ at every coordinate, because an endpoint of $[-1,1]$ cannot be a nontrivial convex average of its points. Thus
$$
\boxed{\operatorname{ext}B_{\ell^\infty}=\{-1,1\}^{\mathbb N}.}
$$
For the <space of sequences converging to zero>, every unit-ball sequence has some coordinate with $|x_j|<1$, since its coordinates tend to zero. The same opposite single-coordinate perturbations still belong to $c_0$, giving
$$
\boxed{\operatorname{ext}B_{c_0}=\varnothing.}
$$
These are <extreme points of real sequence-space unit balls>. There is no conflict with Krein-Milman: the norm-closed unit ball of $c_0$ is not <compact> in the <norm topology>, as the coordinate vectors have mutual distance one.
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