Separability of $X$ and $X^*$ Let $X$ be any normed space, $X^*$ - its dual space (of bounded linear maps $X\longrightarrow\mathbb{C}$). We suppose $X^*$ to be separable and need to prove that $X$ is separable also (i.e. it has a dense countable subset).
Suppose $X$ is not separable. It means that there is an uncountable set $X'=\{x_\alpha\}$ such that for each $\alpha, \alpha'$ we have $||x_\alpha-x_{\alpha'}||\ge1$. So, our goal is to build starting with $X'$ some uncountable set of bounded functionals $X\longrightarrow \mathbb{C}$ with the same property.
So, is it possible to prove the separability of $X$ this way?
We need to build an uncountable set of of linear functionals - for each vector $x_\alpha$ to find some linear functional...
 A: Note first there exists an $M$ such that $\|x_\alpha\| \le M$ for uncountably many $\alpha$.  (Otherwise $X' = \bigcup_{M=1}^\infty \{ x_\alpha : \|x_\alpha\| \le M\}$ is a countable union of countable sets and hence countable.)  So passing to a subset, let us assume $\|x_\alpha\| \le M$ for all $\alpha$. 
Using the axiom of choice, let $\le$ be a well ordering on the index set $A$.  Since $A$ is uncountable, we have $(A, \le) \ge \omega_1$, where $\omega_1$ is the least uncountable ordinal.  So passing to a further subset, let us assume $A = \omega_1$.  We now have $X' = \{ x_\alpha : \alpha \in \omega_1\}$, where $\|x_\alpha\| \le M$ and $\|x_\alpha - x_{\alpha'}\| \ge 1$.
The idea is this: using Hahn-Banach, construct a family $\{f_\alpha : \alpha \in \omega_1\} \subset X^*$ such that $f_\alpha(x_\alpha) = M$ and $f_\alpha(x_\beta) = 0$ for $\beta < \alpha$.  The snag is that this may not be possible due to linear dependence among the $x_\alpha$.  However, we will show this snag cannot happen too often.  
Namely, for each $\alpha < \omega_1$, let $E_\alpha$ be the closed linear span of $\{x_\beta : \beta < \alpha\}$.  Note that $E_\alpha$ is separable.  Now let $B = \{\alpha < \omega_1 : x_\alpha \notin E_\alpha\}$.  If $B$ is countable then it is bounded: there exists $\alpha_0 < \omega_1$ such that $\alpha < \alpha_0$ for all $\alpha \in B$.  This means that for all $\alpha \ge \alpha_0$, we have $x_\alpha \in E_{\alpha}$, i.e. $X' \subset E_{\alpha_0}$.  (Proceed by transfinite induction: let $\alpha \ge \alpha_0$ and suppose for all $\beta < \alpha$ we have $x_\beta \in E_{\alpha_0}$.  Since $E_\alpha$ is the closed linear span of all those $x_\beta$, we have $E_\alpha \subset E_{\alpha_0}$.  But $x_\alpha \in E_\alpha$ since $\alpha \ge \alpha_0$, hence $x_\alpha \in E_{\alpha_0}$.) But $X' \subset E_{\alpha_0}$ is absurd since $E_{\alpha_0}$ is separable and $X'$ is an uncountable set with pairwise distances 1.  So $B$ must be uncountable.
Now for $\alpha \in B$, use Hahn-Banach to find a bounded linear functional $f_\alpha$ with $f_\alpha(E_\alpha) = 0$ and $f_\alpha(x_\alpha) = M$.  I claim $\{f_\alpha : \alpha \in B\}$ is the desired set.  For if $\alpha, \beta \in B$ with $\alpha < \beta$, we have $f_\alpha(x_\alpha) = M$ but $f_\beta(x_\alpha) = 0$.  Since $\|x_\alpha\| \le M$, we have $\|f_\alpha - f_\beta\| \ge 1$.  $B$ is uncountable, so we are done.
