If $X$ is a Banach space, in class we defined the family $\mathcal{P}$ of seminars on $X$ as $$ \mathcal{P} := \left\{ p_{\omega} \mid \omega \in X^{*} \right\} $$ with $p_{\omega} (x) = | \omega(x)|$ for all $x \in X$. This gives the weak topology on $X$.
Problem: Let $X$ be a Banach space whose weak and norm topologies coincide. Prove there exist $\omega_1, \ldots, \omega_n \in X^{*}$ such that $$ || x|| \leq \max_{i=1 \ldots, n} | \omega_i (x)|. $$
Attempt: I started as follows. Let $B_{norm} (0,1) = \left\{ x \in X \mid || x|| < 1 \right\}$ be an open ball centered around $0$ in the norm topology. Since this open set is also weakly open, by definition for every $x \in B_{norm} (0,1)$ there exists a $\delta > 0$ and $\omega_1, \ldots, \omega_n \in X^{*}$ such that $$ B_1 (x, \delta) \cap \ldots \cap B_n (x, \delta) \subset B_{norm} (0,1). $$ Here $B_i(x, \delta) := \left\{ y \in X \mid d_i (x,y) < \delta \right\} = \left\{ y \in X \mid | \omega_i (x-y) | < \delta \right\}$.
However, I'm not sure how to continue. Any help is appreciated.
Such a Banach space $X$ must be finite dimensional, thus if $\{e_1,\dots,e_n\}$ is a normalized basis with coordinate functionals $e_i^*\in X^*$, then
$$||x||=||\sum_{i=1}^n e_i^*(x)e_i||\le \sum_{i=1}^n |e_i^*(x)| \,||e_i||\le n\max_i|e_i^*(x)|$$
Then one can choose $\omega_i^*=ne_i^*$