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Let $X$ be a Banach space.
By Banach-Alaoglu and Krein-Milman Theorems, one can show that if $X$ is a dual space, then $X$ must have at least one extreme point of the closed unit ball.

I am interested in its converse.
More precisely,

Question: Let $X$ be a Banach space.
If the closed unit ball of $X$ has at least one extreme point, must $X$ be a dual space?

I feel that the statement above is negative.
However, I could not produce a counterexample.

In fact, the only Banach spaces which I know that are not dual spaces are $c_0$ and $C_0(mathbbR)$ (the latter set is the collection of all real-valued continuous function vanishing at infinity) because both sets have no extreme point.

Every separable Banach space $X$ can be equivalently renormed so that every point in the unit sphere is an extreme point: Take an injective bounded linear operator $T$ from $X$ into $ell_2$ and use $|x| := |x|_X + |Tx|_2$. Of course, there are many separable Banach spaces that are not isomorphic to a separable conjugate space, including (as Dirk pointed out) those that fail the Radon Nikodym property.

No. Let $X$ and $Y$ be Banach spaces, and set $Z=Xoplus Y$, with $||(x,y)||:=|x|+|y|$. Assume that $x$ is a extreme point of $X$ with $|x|=1$. Then $(x,0)$ becomes an extreme point of $Z$; indeed, if
$$(x,0)=frac12(a,y)+frac12(b,z)$$
for $(a,y),(b,z)$ in the unit ball of $Z$, we then have $a=x=b$, since $x$ is an extreme point, but then $1=|x|=|a|leq||(a,y)||leq 1$, so $y=0$, and analogously, $z=0$.

So, $L^2(mathbb R)oplus L^1(mathbb R)$, is not a dual space, but its unit ball has extreme points.