= Solution
Suppose the weak-star closure of $Z$ were a proper linear <vector subspace> of $X^*$. The <Hahn-Banach separation theorem> would give a nonzero weak-star <continuous linear functional> vanishing on it. A <continuous dual of a weak-star topology> consists precisely of evaluations at points of $X$: <continuity> bounds the <linear functional> by finitely many evaluations, so it factors through their finite-dimensional coordinate map and is a linear combination of them. Thus some $0\neq x\in X$ would satisfy $z(x)=0$ for all $z\in Z$. The norming inequality would imply $c\|x\|\leq0$, a contradiction. \b[Every <norming subspace> of $X^*$ is weak-star dense.]
For the infinite-codimension example, take
$$
\boxed{X=\ell^1(\mathbb N;\mathbb R),\qquad X^*=\ell^\infty(\mathbb N;\mathbb R),\qquad Z=c_0.}
$$
Here $X$ is the <absolutely summable sequence space> and $c_0$ is the <space of sequences converging to zero>, a norm-closed <vector subspace> of $\ell^\infty$. The dual identification is $y(x)=\sum_nx_ny_n$: a bounded sequence defines a <linear functional> of <norm> $\|y\|_\infty$, and every <linear functional> on $\ell^1$ has this form by evaluating on the coordinate vectors and using density of finitely supported sequences.
For $x\in\ell^1$, use the finitely supported sequence $y^{(N)}$ whose first $N$ entries are $\operatorname{sgn}(x_n)$ and whose remaining entries vanish. It belongs to $c_0$, has <norm> at most one, and
$$
y^{(N)}(x)=\sum_{n=1}^N|x_n|\longrightarrow\|x\|_1.
$$
The reverse inequality follows from $|y(x)|\leq\|y\|_\infty\|x\|_1$. Hence \b[$c_0$ is 1-norming for $\ell^1$], an instance of <vanishing sequences norm the summable sequence space>.
To prove infinite codimension, take disjoint infinite sets
$$
A_j=\{2^{j-1}(2m-1):m\geq1\},\qquad j\geq1.
$$
Their indicator sequences have linearly independent classes in $\ell^\infty/c_0$. Indeed, a finite combination has the constant value $a_j$ on $A_j$; if it tends to zero, every $a_j$ must vanish. Thus the quotient is infinite-dimensional.
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