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This seems plausible, given the properties of the unit ball of $c_0$.

I have a compact set in a complex Banach space $X$ whose closed convex hull has uncountably many extreme points. It would be nice to deduce from this that $X$ contains no copy of $c_0$. I have been searching, but could find no proof either way, and I cannot see how to prove it myself---well, not yet... But in the mean time, perhaps this is already known.

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    $\begingroup$ Consider the set $\{(x_1,x_2,x_3,\ldots):x_1^2+x_2^2\le1\text{ and }x_i=0\text{ for }i\ge3\}$. $\endgroup$
    – Yoav Kallus
    Commented Sep 25, 2015 at 0:34

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As stated your question admits an immediate answer because the extreme point structure of finite dimensional convex sets in infinite-dimensional Banach spaces is not related to the structure of the Banach space: for any such set we can find an affine (and thus, preserving extreme structure) map into any other infinite-dimensional Banach space.

Comment of Yoav Kallus is an illustration of this.

On the other hand, there is a very interesting theory of extreme points of unit balls of Banach spaces, in which $c_0$ plays an important role. See, for example, the paper of Fonf on Polyhedral Banach spaces.

Possibly it is worthwhile to redesign your question.

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