On norming weakly$^*$ sequences in the dual of the Banach space $c_0$ A bounded subset $B$ of the dual $X^*$ of a Banach space $X$ is called norming if the formula $\|x\|:=\sup\{|x^*(x)|:x^*\in B\}$ determines an equivalent norm on $X$.
Observe that the sequence $(e_n^*)_{n\in\omega}\subset c_0^*=\ell_1$ of coordinate functionals in the dual of $c_0$ is norming and weakly$^*$ null.

Question 1. Is it true that every absolutely convex bounded norming set $B\subset c_0^*$ in the dual of the Banach space $c_0$ contains a norming weakly$^*$ null sequence?

The same question can be asked for subspaces of $c_0$.

Question 2. Let $X$ be a closed subspace of the Banach space $c_0$ and $B\subset X^*$ is a norming bounded absolutely convex set. Is it true that $B$ contains a norming weakly$^*$ null sequence? 

Remark. If a Banach space $X$ admits a norming weakly$^*$ null sequence of functionals $\{f_n\}_{n\in\omega}\subset X^*$, then the operator $T:X\to c_0$, $T:x\mapsto (f_n(x))_{n\in\omega}$, is an isomorphic embedding of $X$ into $c_0$.
Therefore, $X$ is isomorphic to a subspace of the Banach space $c_0$.
 A: The answer to your Question 1 (and thus Question 2, for general
subspace) is "No".
Example. Let $B=B_{\ell_1}\cap \ker {\mathbf{1}}$, where
$B_{\ell_1}$ is the unit ball of $\ell_1$ and
${\mathbf{1}}=(1,\dots,1,\dots)\in\ell_\infty$ (it is well-known
and easy to check that $B$ is norming).
Suppose that there is a weak$^*$ null norming sequence
$\{x_i^*\}_{i=1}^\infty$ inside $B$. We may assume, slightly
decreasing the norming constant, that
(1) $x_i^*$ are finitely supported.
(2) eventually they get more and more zeros at the beginning.
Let $c$ be the norming constant of this sequence, that is
$\sup_i|x_i^*(x)|\ge c||x||$ for every $x\in c_0$. Let
$\alpha\in(0,1)$ be such that $1-\alpha\le c$. We consider the
following sequence $u$ in $c_0$:
$$u=(1,\underbrace{\alpha,\dots,\alpha}_{n_1}, \underbrace{\alpha^2,\dots,\alpha^2}_{n_2},\alpha^3,\dots),$$
where $n_1$ is such that all $x_i^*$ whose support contains $1$,
are supported in $I_1:=\{1,\dots,n_1+1\}$; $n_2$ is such that all
$x_i^*$ whose support intersects $I_1$ are supported in
$I_2:=\{1,\dots,n_1+n_2+1\}$, so on.
We claim that $|x_i^*(u)|\le\frac{c}2$ for each $i$, thus getting
a contradiction.
In fact, the vector $u$ is such that the support of each $x_i^*$
is contained in the set, where the coordinates of $u$ are either
$\alpha^k$ or $\alpha^{k+1}$ for some $k\in\{0,1,2,3,\dots\}$,
also $||x^*_i||_{\ell_1}\le 1$ and ${\mathbf{1}}(x_i^*)=0$. Thus
$|x_i^*(u)|$ does not exceed the value which we get if the
positive support of $x_i^*$ corresponds to $\alpha^k$ and the
negative support of $x_i^*$ corresponds to $\alpha^{k+1}$, or the
other way around. Therefore $|x^*_i(u)|\le
\frac12(\alpha^k-\alpha^{k+1})\le\frac12(1-\alpha)\le \frac c2$.
