Show that if $X^*$, the dual space of a normed space $X$, is separable, then $X$ is also separable.
My Solution:
Since $X^*$ is separable, there exists a countable dense subset $D \subseteq X^*$. To show that $X$ is separable, we need to find a countable dense subset in $X$.
For each linear functional $f \in D$, by the Hahn-Banach theorem, there exists an $x_f \in$ $X$ with $\left\|x_f\right\|=1$ such that $f\left(x_f\right)=\|f\|$. Define $S=\left\{x_f: f \in D\right\}$. Since $D$ is countable, $S$ is also countable.
To prove that $S$ is dense in $X$, consider any $x \in X$ and $\varepsilon>0$. For every $f \in X^*$, there exists a $g \in D$ such that $\|f-g\|<\varepsilon /(2\|x\|)$ because $D$ is dense in $X^*$. For this $g$, we have $x_g \in S$ where $\left|f\left(x_g\right)-g\left(x_g\right)\right|<\varepsilon$, implying that $S$ is dense in $X$.
Therefore, $X$ is separable, as we have found a countable dense subset $S$ in $X$.
Consider the space $\ell^1$, the space of all absolutely summable sequences. $\ell^1$ is a classic example of a separable space. It has a countable dense subset consisting of sequences with only a finite number of non-zero elements, each of which is a rational number.
However, its dual space, $\left(\ell^1\right)^*$, is isometrically isomorphic to $\ell^{\infty}$, the space of all bounded sequences. This is a standard result in functional analysis. The dual pairing is given by the action of a bounded sequence on an absolutely summable sequence, summing the products of their terms.
The space $\ell^{\infty}$ is known to be non-separable. This can be shown by considering the subset of $\ell^{\infty}$ consisting of all sequences that take only the values 0 and 1 . For any two distinct sequences in this subset, their difference has norm 1, so no countable subset can be dense in $\ell^{\infty}$.
This is my solution, I don't have a specific question but a general proofread and any suggestions would be great :)
You don't say how you use Hahn-Banach for this. But the result is false. Here is an easy example: let $X=\{f\in C[0,1]:\ f(0)=0\}$ with the supremum norm, and let $\varphi(f)=\int_0^1 f$. Then $\|\varphi\|=1$ , but $|\varphi(f)|<1$ for all $f\in X$.
You would have to explain why having $|f(x_g)-g(x_g)|<\varepsilon$ somehow implies that $x$ is close to $x_g$. Note also that all elements in your $S$ have norm equal to $1$, so $S$ cannot possibly be dense. You might be able to show that $S$ is dense in the unit sphere and follow from there, but even then your argument has the problem that such $S$ doesn't exist.