This is my first time posting here. I already saw a post regarding this result, but I wanted to check if this specific proof is correct or not; as the proof presented in the book is different. Also, I'm not a native English speaker, so excuse me if there's any grammar mistakes.
Proposition: Let $(E,||.||)$ be a normed space (over $\mathbb{R}$ or $\mathbb{C}$) such that $E^{*}$ is separable. Then $E$ is separable as well.
Proof: Let $D = \{\phi_{n}\}_{n \in \mathbb{N}}$ be a dense in $E^{*}$.
Define $x_{n} \in E$ such that $||x_{n}||=1$ and $||\phi_{n}|| < |\phi_{n}(x_{n})|+\frac{1}{n}$, which exists by definition of $||\phi_{n}||$. I claim that $\langle x_{1},...x_{k}...\rangle$ is a dense in $E$. If I could prove that, due to a previous result, this would imply that $E$ is separable. We'll define $S=\overline{\langle x_{1},...x_{k}...\rangle}$. We need to prove that $S=E$.
Let's assume otherwise. Then there would exist $x \in E-S$, and given that $S$ is a subspace of $E$ we can assume $||x||=1$. By a corollary of the Hahn-Banach theorem (since $S$ is closed), we can find a continious functional $\phi \in E^{*}$ such that $\phi(S)=\{0\}$ and $\phi(x)=1$.
Now, we know that there exists a sequence in $D$ such that $\phi_{n_{k}} \rightarrow \phi$, but if that's the case:
$$||\phi_{n_{k}}||-\frac{1}{n_{k}}<|\phi_{n_{k}}(x_{n_{k}})|=|\phi_{n_{k}}(x_{n_{k}})-0|=|\phi_{n_{k}}(x_{n_{k}})-\phi(x_{n_{k}})|\leq ||\phi_{n_{k}}-\phi||,$$
which implies that $||\phi_{n_{k}}|| \rightarrow 0$ as $k \rightarrow \infty$. For $k$ large enough, $|\phi_{n_{k}}(y)| < \frac{1}{2}$ for all $y$ with $||y||=1$, in particular, $|\phi_{n_{k}}(x)| < \frac{1}{2}$. Then,
$$||\phi-\phi_{n_{k}}|| \geq |\phi(x)-\phi_{n_{k}}(x)| = |1-\phi_{n_{k}}(x)| >\frac{1}{2}$$
But $||\phi-\phi_{n_{k}}|| \rightarrow 0$? We reached our contradiction and conclude that $S=E$.
Thank you in advance.