As noted by kimchi, if there is $i \in \mathbb{N}$ with $|x_i| < 1$ then $x$ is not an extreme point.
Now consider $x \in c$ with absolute value of all of its coordinates equal to $1$. Suppose $x = \frac{1}{2} y + \frac{1}{2} z$ for some $y,z \in c$. For arbitrary $n \in \mathbb{N}$ we have $|x_n| = 1$ hence $x_n$ in an extreme point of the unit ball $B_{\mathbb{C}}$ of $\mathbb{C}$. As $y_n,z_n \in B_{\mathbb{C}}$ and $x_n = \frac{1}{2}y_n + \frac{1}{2}z_n$, we get $x_n =y_n=z_n$. Now $n\in \mathbb{N}$ was arbitrary, hence, $x=y=z$ and $x$ is an extreme point.
To sum up the extreme points of $c$ are exactly the convergent sequences of complex units.
As noted in the comments, for the two Banach spaces $c_0$ and $L^1[0,1]$, the unit ball has no extreme points.
Here is $L^1[0,1]$.
Write $\|\cdot\|$ for the $L^1$ norm. Write $B := \{f \in L_1 : \|f\| \le 1\}$ for the unit ball. Let $h \in B$. We claim
$h \not\in \operatorname{Ext}(B)$.
Case 1. $h = 0$. Then $h = \frac12(\mathbf1 + (-\mathbf1))$ where $\mathbf1$ is the constant function with value $1$. Since $\mathbf1 \in B$ and $-\mathbf1 \in B$ and $\mathbf1 \ne -\mathbf1$, we conclude that $h$ is not an extreme point of $B$.
Case 2. $0 < \|h\| \le 1$. Define the function $\phi : [0,1] \to \mathbb R$ by
$$
\phi(t) = \int_0^t |h|
$$
Then $\phi$ is continuous, $\phi(0) = 0$ and $\phi(1) = \|h\| > 0$.
so there is $t_0 \in (0,1)$ so that $\phi(t_0) = \frac12 \|h\|$. Now
$h = \frac12(h_1+h_2)$ where
$$
h_1 = 2 \mathbf1_{[0,t_0]} h, \qquad h_2 = 2 \mathbf1_{(t_0,1]} h
$$
Then $\|h_1\| = \|h_2\| = \|h\| \le 1$ so $h_1, h_2 \in B$. Also, $h_1 \ne h_2$. So we again conclude that $h$ is not an extreme point of $B$.
You can find a lot more about this interesting topic in
Diestel, J.; Uhl, J. J., Vector measures, Mathematical Surveys. No. 15. Providence, R.I.: American Mathematical Society (AMS). XIII, 322 p. $ 35.60 (1977). ZBL0369.46039.
Best Answer
No. The unit ball of the space $c_0$ of all null sequences does not have any extreme points.