If $E$ is separable, why can we find $(x'_n)\subseteq E'$ such that $\langle x,x'_n\rangle$ for all $n$ implies $x=0$?

Question

Let $E$ be a separable normed $\mathbb R$-vector space.

If $x\in E\setminus\{0\}$, by the Hahn-Banach theorem, then there is a $x'\in E'$ with $\left\|x'\right\|_{E'}=1$ and $\langle x,x'\rangle=\left\|x\right\|_E$.

How can we show that there is a $(x'_n)_{n\in\mathbb N}\subseteq E'$ such that $\langle x,x'_n\rangle =0$ for all $n\in\mathbb N$ implies $x=0$?

Since $E$ is separable, there is a dense $(x_n)_{n\in\mathbb N}\subseteq E\setminus\{0\}$ and as mentioned above, we can find $(x'_n)_{n\in\mathbb N}\subseteq E'$ with $\left\|x'_n\right\|_{E'}=1$ and $\langle x_n,x'_n\rangle=\left\|x_n\right\|_E$ for all $n\in\mathbb N$.

Now let $x\in E$ with $\langle x,x'_n\rangle=0$ for all $n\in\mathbb N$. Since $(x_n)_{n\in\mathbb N}$ is dense, there is an increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$ with $\left\|x_{n_k}-x\right\|_E\xrightarrow{k\to\infty}0$; but how can we conclude?

Answer 1

We have $$||x_{n_k}||_E=|x_{n_k}'(x_{n_k}) -x_{n_k}'(x)|\leq ||x_{n_k} - x||_E\to 0$$ Thus $x_{n_k} \to x $ and $x_{n_k} \to 0$ therefore $x=0.$