Why a closed bounded convex set in $\mathbb{R}^{n}$ always has an extreme point? Let $X\subseteq \mathbb{R}^{n}$ is closed, bounded convex set. How to prove that $X$ contains such point $x$ that we can't represent as $x=\frac{1}{2}x_{1}+\frac{1}{2}x_{2}$ where $x_1\in X$ and $x_2\in X$ and $x_1 \neq x_2$?
For example closed square in $\mathbb{R}^{2}$ has $4$ such points.
 A: Let $R$ be the diameter of $X$. Choose any $x, y \in X$ such that $\lVert x - y\rVert = R$; prove by contradiction that $x$ is the desired point.
Basically, suppose that $x = \frac{1}{2}x_1 + \frac{1}{2}x_2$ for some distinct $x_1, x_2 \in X$. Note that either $\lVert x_1 - y\rVert$ or $\lVert x_2 - y\rVert$ must exceed $R$.
Detailed Proof:
Suppose that $x = \frac{1}{2}x_1 + \frac{1}{2}x_2$ for some distinct $x_1, x_2 \in X$. Let $x_1 = x + v$ and $x_2 = x - v$. Since $x_1 \neq x_2$, $v \neq 0$.
$$
\begin{align}
\lVert x_2 - y \rVert &\leq R\\
\implies \lVert x_2 - y \rVert^2 &= ((x - y) - v) \cdot ((x - y) - v)\\
                                 &= R^2 + \lVert v \rVert^2 - 2v \cdot (x - y)\\
                                 &\leq R^2
\end{align}
$$
Hence, $2v \cdot (x - y) \geq \lVert v \rVert^2 \geq 0$ and thus $R^2 + \lVert v \rVert^2 + 2v \cdot (x - y) \geq R^2$ ($\ast$).
Similarly,
$$
\begin{align}
\lVert x_1 - y \rVert &\leq R\\
\implies \lVert x_1 - y \rVert^2 &= ((x - y) + v) \cdot ((x - y) + v)\\
                                 &= R^2 + \lVert v \rVert^2 + 2v \cdot (x - y)\\
                                 &\leq R^2
\end{align}
$$
Together with ($\ast$), this gives us $R^2 + \lVert v \rVert^2 + 2v \cdot (x - y) = R^2$, which implies that $\lVert x_2 - y \rVert = \lVert x_2 - y \rVert = R$.
Here comes the part where the Euclidean metric is crucial. Since $x_1$ and $x_2$ lie on the sphere of radius $R$ and centered at $y$, $x$ is the midpoint of the chord joining $x_1$ and $x_2$, so $x - y$ is perpendicular to $v$. The Pythagorean Theorem gives us $R^2 = \lVert x - y\rVert^2 = R^2 - \lVert v\rVert^2 < R^2$. Contradiction!
