Let $X$ be a normed separable space. Prove that there is a sequence $x^*_n\in X^*$ that satisfy: $\forall 0\neq x\in X$ there is infinite $n$'s such that: $x^*_nx\neq 0$.
By a corollary of Hahn Banach which states: If $X$ is a normed space, $x\neq0 \in X$ then there's $x^*\in X^*$ such that $||x^*||=1$ and $x^*x=||x||$.
Since $X$ is separable so I can assume that there is infinite non-zero $x\in X$ (a non-zero sequence $x_n$ in $X$) so for each such $x$ there is $x^*_i\in X^*$ such that $||x^*_i||=1$ and $x^*_ix=||x||$.Since $x\neq 0$ then $x^*_ix=||x||\neq 0$.
I am not sure if was enough coear. Any improvements would be welcomed.
Let $S$ be a countable dense subset in $X$ and for every $x \in S$ a corollary of Hahn-Banach gives a functional $x^* \in X^*$ such that $\|x^*\|=1$ and $x^*(x) = \|x\|$. I claim that $\{x^* : x \in S\}$ is your desired countable family of functionals.
Let $x \in X$, $x \ne 0$ be arbitrary. There exists a sequence $(x_n)_n$ in $S$ such that $x_n \to x$. We have
$$\|x_n\|-x_n^*(x)\le |x_n^*(x_n) - x_n^*(x)| = |x_n^*(x_n-x)| \le \|x_n^*\|\|x_n-x\| = \|x_n-x\|$$ so $$x_n^*(x) \ge \|x_n\|-\|x_n-x\|.$$ I claim that for infinitely many $n\in\Bbb{N}$ we have $\|x_n\|-\|x_n-x\| > 0$. Otherwise there exists some $n_0\in\Bbb{N}$ such that for $n \ge n_0$ we have $$\|x_n\| \le \|x_n-x\| \to 0$$ so $x_n \to 0$. This is a contradiction since $x_n \to x$ and $x \ne 0$.
Therefore $x_n^*(x) > 0$ for infinitely many $n\in\Bbb{N}$.