[proofplan]
This is a direct application of the [Krein-Milman Theorem](/theorems/2661) to the dual unit ball equipped with the weak* topology. We verify the three hypotheses required by Krein-Milman: $B_{X^*}$ is convex, the weak* topology makes $X^*$ a locally convex space, and $B_{X^*}$ is weak*-compact. The first follows from the triangle inequality, the second from the construction of the weak* topology as initial topology with respect to a separating family of seminorms, and the third from the **Banach-Alaoglu Theorem**. The conclusion is immediate.
[/proofplan]
[step:Verify that $B_{X^*}$ is convex]
Let $f, g \in B_{X^*}$ and $t \in [0, 1]$. By the triangle inequality and homogeneity of the operator norm,
\begin{align*}
\|tf + (1-t)g\|_{X^*} \leq t\|f\|_{X^*} + (1-t)\|g\|_{X^*} \leq t \cdot 1 + (1-t) \cdot 1 = 1.
\end{align*}
Hence $tf + (1-t)g \in B_{X^*}$, and $B_{X^*}$ is convex.
[/step]
[step:Verify that $(X^*, w^*)$ is a locally convex space]
The weak* topology on $X^*$ is by definition the initial topology induced by the family of evaluation maps $\{\hat{x} : x \in X\}$, where
\begin{align*}
\hat{x}: X^* &\to \mathbb{K} \\
f &\mapsto f(x).
\end{align*}
Equivalently, $w^*$ is the topology generated by the family of seminorms $\{p_x\}_{x \in X}$, where $p_x(f) := |f(x)|$.
Each $p_x$ is a seminorm: $p_x(f + g) = |f(x) + g(x)| \leq |f(x)| + |g(x)| = p_x(f) + p_x(g)$ and $p_x(\lambda f) = |\lambda f(x)| = |\lambda| p_x(f)$.
The family $\{p_x\}_{x \in X}$ separates points of $X^*$: if $f \neq g$, there exists $x \in X$ with $f(x) \neq g(x)$, hence $p_x(f - g) = |f(x) - g(x)| > 0$.
A topology generated by a separating family of seminorms is locally convex Hausdorff. Therefore $(X^*, w^*)$ is a locally convex space.
[/step]
[step:Verify that $B_{X^*}$ is weak*-compact via Banach-Alaoglu]
By the **Banach-Alaoglu Theorem**, the closed unit ball $B_{X^*}$ of the dual of a normed space is compact in the weak* topology.
[guided]
The **Banach-Alaoglu Theorem** states: for any normed space $X$, the closed dual unit ball $B_{X^*} = \{f \in X^* : \|f\|_{X^*} \leq 1\}$ is compact in the weak* topology $w^* = \sigma(X^*, X)$.
The hypotheses are: $X$ is a normed space (so $X^*$ has a well-defined operator norm and $B_{X^*}$ is well-defined). This is given.
The proof of Banach-Alaoglu (recalled here as motivation) embeds $B_{X^*}$ into the product space $\prod_{x \in X} \overline{B}(0, \|x\|) \subset \mathbb{K}^X$ via $f \mapsto (f(x))_{x \in X}$. Each factor $\overline{B}(0, \|x\|)$ is compact in $\mathbb{K}$, so the product is compact by **Tychonoff's Theorem**. The image of $B_{X^*}$ under the embedding is closed in this product (defined by linearity equations $\pi_{x+y} = \pi_x + \pi_y$ and $\pi_{\lambda x} = \lambda \pi_x$, all of which are closed conditions), hence compact. The weak* topology on $B_{X^*}$ is precisely the subspace topology pulled back through this embedding, so $(B_{X^*}, w^*)$ is compact.
Therefore $B_{X^*}$ is weak*-compact, completing the verification of the Krein-Milman hypothesis.
[/guided]
[/step]
[step:Apply Krein-Milman to conclude]
We have verified:
1. $B_{X^*}$ is a non-empty (it contains the zero functional $0 \in X^*$ with $\|0\|_{X^*} = 0 \leq 1$) convex subset of $X^*$.
2. $(X^*, w^*)$ is a locally convex space.
3. $B_{X^*}$ is compact in $w^*$.
By the [Krein-Milman Theorem](/theorems/2661) applied to the compact convex non-empty set $B_{X^*}$ inside the locally convex space $(X^*, w^*)$,
\begin{align*}
B_{X^*} = \overline{\operatorname{conv}}^{w^*}(\operatorname{Ext}(B_{X^*})).
\end{align*}
In particular, since $B_{X^*} \neq \varnothing$, the "in particular" conclusion of Krein-Milman gives $\operatorname{Ext}(B_{X^*}) \neq \varnothing$.
[/step]
