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I'm reading @Nate Eldredge's answer:

Here's a counterexample. Let $X$ be any infinite-dimensional Banach space.

Let $\mathcal{U}$ be the collection of all weak-* open neighborhoods of $0 \in X^*$. One can show that each $U \in \mathcal{U}$ contains a line. (Think about what the basic open sets are. Indeed, $U$ contains a vector subspace of finite codimension.) For each $U$, let $f_U$ be a nonzero point on such a line, so that $\mathbb{R} f_U \subset U$.

Let $U \in \mathcal U$. There are $x_1, \ldots, x_n \in X$ and $\varepsilon>0$ such that $$ U \supset V := \{f \in X^* \mid |\langle f , x_i \rangle| <\varepsilon \text{ for all } i =1,\ldots,n \}. $$

Clearly, $f\in V \iff -f \in V$. Now we fix $f_0 \in V$. Then $\lambda_i := |\langle f_0 , x_i \rangle|$ for all $i =1,\ldots,n$. Let $\lambda := \min_i (\varepsilon/\lambda_i) >1$. Then $rf_0 \in V$ for all $r\in (-\lambda, \lambda)$.

What I got is a segment $(-\lambda, \lambda)$. Could you explain what the line is in this context?

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    By Hahn-Banach there is some $f_0 \in X^\ast\setminus {0}$ such that $f_0(x_i)=0$ $(i=1,\dots,n)$. Then $\lambda f_0 \in V$ for all $\lambda$. – Gerd Feb 20 '22 at 11:13

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From @Gerd's hint, I give an explicit argument below.

Consider $T:X^* \to \mathbb R^n, f \mapsto (f(x_1), \ldots, f(x_n))$. Then $T$ is linear and continuous in weak* topology. If $\operatorname{ker} T = \{0\}$, then $T$ is injective and thus $\dim X^* \le n$. This in turn implies $\dim X = n$, which is a contradiction. Hence, there is $0 \neq f \in E^*$ such that $f(x_i)=0$ for all $i=1, \ldots, n$.

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