= Solution
An <extreme point> $x$ of a <convex set> $K$ is one for which $x=(1-t)y+tz$, $y,z\in K$ and $0<t<1$ imply $y=z=x$. It is enough to test midpoint decompositions: a nontrivial interior point of a segment is itself the midpoint of a shorter nontrivial segment contained in that segment.
We prove the <Krein-Milman theorem>. Assume first that $K$ is nonempty. A <face of a convex set> $K$ means a convex subset $F$ with the property that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. The family of nonempty compact faces of $K$, including $K$ itself, has a minimal member by the <Zorn lemma>: every inclusion-decreasing chain has a nonempty intersection by <compactness> and the <finite intersection property>, and that intersection is again a compact face.
Let $M$ be such a minimal face. If it contains two distinct points, the continuous dual of a Hausdorff <locally convex space> separates them, by the permitted <Hahn-Banach theorem>. Choose a continuous real <linear functional> $\ell$ that is not constant on $M$. It attains a maximum on the compact set $M$, and
$$
M_1=\{z\in M:\ell(z)=\max_M\ell\}
$$
is a nonempty proper compact face of $M$, and hence a face of $K$. Indeed, equality at the maximum in a convex combination forces both endpoint values to be maximal. This contradicts minimality. Thus $M$ is a singleton, and its point is extreme in $K$. The same argument applied within any nonempty compact face proves that every such face contains an <extreme point> of $K$.
Let $C=\overline{\operatorname{conv}}(\operatorname{ext}K)$. It is contained in $K$, since the latter is closed and convex, and <extreme points> exist when $K$ is nonempty. If $x_0\in K\setminus C$, the permitted separation theorem provides a continuous real <linear functional> $\ell$ with $\ell(x_0)>\sup_C\ell$. Let $F$ be its maximizer face on $K$. This nonempty compact face contains an <extreme point> $v$ of $K$, but
$$
\ell(v)=\max_K\ell\geq\ell(x_0)>\sup_C\ell,
$$
contradicting $v\in\operatorname{ext}K\subseteq C$. Therefore
$$
\boxed{K=\overline{\operatorname{conv}}(\operatorname{ext}K)}.
$$
For empty $K$, the equality holds with the usual empty convex-hull convention. This proves both the extreme-point existence step and the closed-convex-hull step; no <compactness> of the set of <extreme points> itself is assumed.
The real <absolutely summable sequence space> is $\ell^1=\{x=(x_j):\sum_j|x_j|<\infty\}$ with <norm> $\|x\|_1=\sum_j|x_j|$. The real <l-infinity sequence space> is $\ell^\infty=\{x=(x_j):\sup_j|x_j|<\infty\}$ with <norm> $\|x\|_\infty=\sup_j|x_j|$. Their <completeness> is allowed as given; their <extreme points> still require proof.
For $\ell^1$, an interior point of the <unit ball> is not extreme because small opposite perturbations in one coordinate stay in the ball. At a boundary point with two nonzero coordinates $i,j$, choose $0<\varepsilon<\min(|x_i|,|x_j|)$ and put
$$
z=\varepsilon\operatorname{sgn}(x_i)e_i-\varepsilon\operatorname{sgn}(x_j)e_j.
$$
The two distinct vectors $x+z,x-z$ both have <norm> one and average to $x$, so $x$ is not extreme. A remaining boundary point must be $\pm e_j$. If $e_j=(y+z)/2$ with $\|y\|_1,\|z\|_1\leq1$, the $j$th coordinate forces $y_j=z_j=1$, exhausting both <norm> budgets and forcing all other coordinates to zero. The negative case is identical. Thus
$$
\boxed{\operatorname{ext}B_{\ell^1}=\{e_j,-e_j:j\geq1\}}.
$$
For $\ell^\infty$, if $|x_j|<1$ in even one coordinate, small opposite changes to that coordinate give a nontrivial midpoint decomposition within the ball. If every $x_j=\pm1$, a midpoint decomposition forces both summands to have that same value in every coordinate, so they equal $x$. Hence
$$
\boxed{\operatorname{ext}B_{\ell^\infty}=\{-1,1\}^{\mathbb N}}.
$$
These are precisely the <extreme points of real sequence-space unit balls> for the two spaces under consideration.
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