= Solution
For each $n$, define $Jx_n:X^*\to\mathbb R$ by $Jx_n(f)=f(x_n)$. Weak convergence makes $(Jx_n(f))_n$ bounded for every $f\in X^*$. The <Uniform boundedness principle> gives
$$
\sup_n\lVert Jx_n\rVert=\sup_n\lVert x_n\rVert<\infty.
$$
Let $K$ be the closed convex hull of the $x_n$. Given a sequence in $K$, approximate its terms in norm by finite convex combinations of the $x_n$. A diagonal subsequence makes every coefficient converge. Any loss of total coefficient mass is assigned to zero, which belongs to $K$ by <Mazur theorem> because $x_n\rightharpoonup0$. Since $f(x_n)\to0$ for every $f\in X^*$, splitting each sum into a finite head and a uniformly small tail proves weak convergence of this subsequence to the corresponding convex combination. Hence $K$ is weakly sequentially compact and, by the stated theorem, weakly compact.
Define
$$
T:X^*\to c_0,
\qquad
Tf=(f(x_n))_{n\geq1}.
$$
It is bounded because $(x_n)$ is norm bounded, and its values lie in $c_0$ because $x_n\rightharpoonup0$. Its adjoint-on-preduals map is
$$
T_*:\ell^1\to X,
\qquad
T_*(a)=\sum_na_nx_n.
$$
If $K$ had nonempty norm interior, then $K-K$ would contain a ball about zero. The quantitative open-mapping argument applied to the convex combinations above would make $T_*$ surjective, and hence make $T$ bounded below. Its range $R$ would be a closed infinite-dimensional subspace of $c_0$ whose unit ball is compact for coordinatewise convergence, since it lies in the coordinatewise compact image of a weak-star compact ball of $X^*$.
By the stated structural theorem, $R$ contains a closed subspace isomorphic to $c_0$. The bounded partial sums of the image of the standard $c_0$ basis would then have a coordinatewise convergent subnet. Uniform boundedness turns coordinatewise convergence in $c_0$ into weak convergence, and norm-closed subspaces are weakly closed. Pulling the limit back would make the partial sums of the standard basis converge weakly in $c_0$, impossible because their coordinate values force the putative limit to be the constant-one sequence. This contradiction proves that $K$ has empty norm interior.
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