Separability of a normed space E I'm stucked with the following problem:
Let $E$ a normed space. Show that $S=\{x\in E\hspace{0.1cm}|\hspace{0.1cm} ||x||=1\}$ is separable if and only if, $E$ is separable.
Could be used the Hann-Banach theorem? and where? I don't see this clearly.
Thanks !
 A: Suppose $S$ is separable, with countable dense subset $S'\subseteq S$. Then, for each $s \in S$ and $\epsilon > 0$, there exists $s' \in S'$ such that $\|s-s'\| < \epsilon$. Let $E'=\{ \rho s' : \rho \in \mathbb{Q}, s' \in S\}=\bigcup_{\rho \in\mathbb{Q},\;\rho > 0}\rho S'$ where $\rho S'=\{ \rho s' : s' \in S\}$. $E'$ is countable because it is the countable union of countable sets. To see that $E'$ is dense in $E$, let $x \in E\setminus\{0\}$ be fixed, and let $\epsilon > 0$ be given. Then there exists $\rho > 0$ in $\mathbb{Q}$ such that $\|x\| < \rho < \|x\|+\epsilon/2$. And there exists $s'\in S$ such that $\|\frac{1}{\|x\|}x -s'\| < \frac{\epsilon}{2\rho}$. So,
$$
\begin{align}
    \|x-\rho s'\| & =\|x\|\| \frac{1}{\|x\|}x-\frac{\rho}{\|x\|}s'\|\\
       & < \|x\|\|\frac{1}{\|x\|}x-s'\|+\|x\|\|s'-\frac{\rho}{\|x\|}s'\|\\
       & < \frac{\epsilon\|x\|}{2\rho}+\|x\|(\frac{\rho}{\|x\|}-1) \\
       & < \frac{\epsilon}{2} +(\rho-\|x\|) \\
       & < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.
\end{align}
$$
Therefore, if $S$ is separable, then $E$ is separable. To show that $S$ is separable if $E$ is separable, let $\{ e_{n} \}$ be a countable dense subset of $E$, discarding $0$ if it is in this set. To show that $\{ \frac{1}{\|e_{n}\|}e_{n}\}$ is a dense subset of $S$, let $s \in S$ and $0 < \epsilon < 1$ be given. Then there exists $k$ such that $\|s-e_{k}\| < \epsilon/2$, which guarantees $e_{k} \ne 0$ with $|1-\|e_{k}\||=|\|s\|-\|e_{k}\||<\|s-e_{k}\|<\epsilon/2$. Hence,
$$
\begin{align}
      \|s - \frac{1}{\|e_k\|}e_k\| & \le \|s-e_k\|+\|e_k-\frac{1}{\|e_k\|}e_k\| \\
        & \le \frac{\epsilon}{2}+\|e_k\||1-1/\|e_k\|| \\
        & = \frac{\epsilon}{2}+|1-\|e_{k}\|| \\
        & < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.
\end{align}
$$
