Proof check about showing unit sphere is separable

Question

I'm trying to prove that if dual space a normed space $X^*$ is separable then $X$ is separable. The first step is to show the closed unit sphere $D^{*}=\left\{\mu \in X^{*}:\|\mu\|=1\right\} $is separable.

My proof is: Let $\{x_n\}_{n \in \mathbb{N}}$ be a dense subset of $X^*$ and let $x \in X^*$ and $\|{x}\|=1$. Then $x$ is arbitrary element in $D^*$ and there exists $n\in \mathbb{N}$ such that $\|{x-x_n}\|<\epsilon$ $\forall \epsilon>0$. Since $\left\|x_{n}\right\| \text { lies between }\|x\|-\epsilon \text { and } \|x\|+\epsilon$ and $\|{x}\|-\|{x-x_n}\| \leq \|{x_n}\| \leq \|{x}\|+ \|{x-x_n}\|$, $1-\epsilon<\left\|x_{n}\right\|<1+\epsilon$.

Since $$x_{n}-\frac{x_n}{\|{x_n}\|}=\left(1-\left\|x_{n}\right\|^{-1}\right) x_{n} $$ we have

$$\left\|\|{x_{n}\|-\frac{x_n}{\|{x_n}\|}}\right\|=\left|1-\left\|x_{n}\right\|^{-1}\right| \cdot\left\|x_{n}\right\|=|1-\| x_{n} \| |$$ Now, $\left\|x-\frac{x_{n}}{\left\|x_{n}\right\|}\right\| \leq\left\|x-x_{n}\right\|+\left\|x_{n}-\frac{x_{n}}{\left\|x_{n}\right\|}\right\|<\epsilon+\left|1-\left\|x_{n}\right\|\right|<2 \epsilon$. Let $\lambda_n=\frac{x_n}{\|{x_n}\|}$, then $\operatorname{set}\left\{\lambda_{n}\right\}_{n \in \mathbb{N}} $ is in $X^*$ and it's dense in $D^*$.

Is there anything wrong? Thanks.

Answer 1

There's nothing wrong with the logic, but I want to highlight the following sentence:

Since $\left\|x_{n}\right\|$ lies between $\|x\|-\epsilon$ and $\|x\|+\epsilon$ and $\|{x}\|-\|{x-x_n}\| \leq \|{x_n}\| \leq \|{x}\|+ \|{x-x_n}\|$, $1-\epsilon<\left\|x_{n}\right\|<1+\epsilon$.

This sentence is a little confusing and a little redundant. It's not immediately clear which statement implies what. I would say something along the lines of

Using the triangle inequality, $$\|x\| - \|x - x_n\| \le \|x_n\| \le \|x\| + \|x - x_n\|.$$ Since $\|x\| = 1$ and $\|x - x_n\| < \varepsilon$, this implies $$1 - \varepsilon < \|x_n\| \le 1 + \varepsilon \implies \Big|1 - \|x_n\|\Big| < \varepsilon.$$