Let $E$ be a Banach space and let $\overline{B}_{E^{*}}$ be the closed unit ball of $E^*$.
$D\subset \overline{B}_{E^{*}}$ is called $\alpha$-norming if $\sup\limits_{v\in D}|\left<e,v\right>|\ge\alpha\|e\|$, for every $e\in E$.
A linear subspace $F\subset E^{*}$ is called $\alpha$-norming if $F\cap \overline{B}_{E^{*}}$ is $\alpha$-norming.
$D$ or $F$ are called norming if they are $\alpha$-norming, for some $\alpha>0$.
Let $D\subset \overline{B}_{E^{*}}$ be balanced convex, and such that that $span D=\cup nD$ is norming, does it follows that $D$ is norming (possibly for different $\alpha$)?
Recall that a set $S$ in a vector space $V$ is called balanced set (or circled, or a disk) if $\alpha S\subset S$, for any scalar $\alpha$ with $|\alpha|\le 1$.
If $\operatorname{span} D$ is norming, it need not be the case that $D$ is norming.
Let $E = \ell^1(\mathbb{N})$, and identify $E^{\ast}$ with $\ell^{\infty}(\mathbb{N})$ as usual. Let $F = c_{00}(\mathbb{N})$ the subspace of sequences with only finitely many nonzero terms, and
$$D = \{\xi \in F : (\forall n \in \mathbb{N})(\lvert \xi_n\rvert \leqslant 2^{-n})\}\,.$$
Then $F$ is a $1$-norming subspace, $F = \operatorname{span} D$, $D$ is a convex and balanced subset of $\overline{B}_{E^{\ast}}$, but $D$ is not norming since
$$\sup_{v \in D}\; \lvert\langle v, e_k\rangle\rvert = 2^{-k}\lVert e_k\rVert$$
for the "standard basis" vectors $e_k$ of $\ell^1(\mathbb{N})$.
The same example (except that one needs to scale $D$ for it to be contained in the closed unit ball) works for $E = \ell^p(\mathbb{N})$ for $1 < p < \infty$, and a similar construction can be carried out for $E = \ell^p(T)$ where $1 \leqslant p < \infty$ and $T$ is an arbitrary infinite set.