I'm new here, so please tell me if there is any mistake.
I study Ch.3 of Sequences and Series in Banach Space by J.Diestel,
and show "Every separable Banach space contains a countable total set"
Def) For given Banach space $X$, a subset $F$ of the dual $X^*$ of $X$ is called total if the following is true : $\begin{equation} f(x)=0 \quad \forall x \in F \quad \Rightarrow x=0. \end{equation}$
The proof follows as : Choose a countable dense subset {$d_n$} of $B_X=\{x \in X: \Vert{x}\Vert=1\}$ and pick $d_n^*$ $\in$ $X^*$, a dual of $X$, such that $\Vert d_n^* \Vert =1$ and $d_n^*(d_n)=1$ for each $n$ (possible by Hahn-Banach Thm). Then {$d_n^*$} is total.
I tried : If $x \neq 0$, there is a subsequence {$d_{n_k}$} of {$d_n$} converging to $x$ and evaluate $d_{n_k}(x)$, but I'm stuck here. Am I going on a right way?
Assume that $x\neq 0$ is such that for each $n$, $d^*_n(x)=0$. We may assume that its norm is equal to $1$.
For any positive $\epsilon$, there exists an integer $N$ such that $\lVert x-d_N\rVert\lt\varepsilon$ (since $\{d_n,n\geqslant 1\}$ is dense in the unit sphere) . Since $d_N^*$ has norm $1$ we have $$\lvert d_N^*\left(x-d_N\right)\rvert\leqslant \left\lVert x-d_N\right\rVert\lt \epsilon.$$ Note that $$d_N^*\left(x-d_N\right)=-1$$ hence we get a contradiction for $\epsilon\leqslant 1$.