extreme points of unit ball in a normed space [duplicate]

Hint: Let $\|x\| < 1$. Then $\overline{B}_r(x) \subset B_X$ with $r := 1 - \|x\|$.

Hence, for any fixed $\|\xi\| = 1$ the points $y := x + r\xi$ and $z:= x- r\xi$ belong to $B_X$, and $$ x = \frac{y + z}{2}\,. $$

Consider making $y $ and $z $ colinear with $x $, make $y $ a bit closer to the "edge" of the ball and $z $ a bit further away, so that $x $ is the midpoint.

In particular, if $\|x\|_X=\rho \lt 1$, and assuming also $\rho\gt 0$ (the case $\rho=0$ is easy), you can define unit vector $u=\frac {1}{\rho} \cdot x $. Take $y=(\rho+\varepsilon)\cdot u $ and $z=(\rho-\varepsilon)\cdot u $ choosing a small $\varepsilon $ and it is easy to prove that $\frac {1}{2}(y+z)=x $ and that $\|y\|_X=\rho+\varepsilon$ and $\|z\|_X=\rho-\varepsilon$ - both norms at most 1 with small enough $\varepsilon $.