Let $X=C[0,1]$ equipped with the maximum norm and define a bounded linear functional $\phi_x(f)=f(x)$ for $x\in [0,1]$. I want to show that $X^*$ is not separable.
There was a hint in this exercise, which told me to compute $\|\phi_x-\phi_y\|$ for $x\neq y$. I was able to show that $\|\phi_x-\phi_y\|=2$ for $x\neq y$ in $[0,1]$. Now I assumed that $X^*$ is separable. I.e that there exists a countable dense subset, say $S\subset X^*$ countable such that $\overline{S}=X^*$. Since for all $x\in [0,1]$ $\phi_x\in X^*$ we know that there exists a sequence $(f_{x,n})_n$ such that $$f_{x,n}\stackrel{n\rightarrow \infty}{\longrightarrow} \phi_x$$
But here I don't now how to continue, I think I should get a contradiction with $\|\phi_x-\phi_y\|=2$. I therefore thought that $$2=\|\phi_x-\phi_y\|\le \|\phi_x-f_{x,n}\|+\|f_{x,n}-f_{y,n}\|+\|f_{y,n}-\phi_y\|$$ the problem here is that I know that $ \|\phi_x-f_{x,n}\|\rightarrow 0$ and $\|f_{y,n}-\phi_y\|\rightarrow 0$ but I have no control on the middle term. Therefore this does not work. Is there another way to get a contradiction?
You can do the following:
Let $S\subseteq (C[0,1])^*$ be a dense subset.
Let $\varpi\colon [0,1]\to S$ be given as follows:
Because $S$ is dense, for each $r\in[0,1]$ there is a $s\in S$ such that $\|\phi_r-s\|<1$. Now let $\varpi(r)$ be such an $s$ (using axiom of choice).
Now we can show that $\varpi$ is injective: If $\varpi(a)=\varpi(b)$, then $\|\phi_a-\phi_b\|\leq\|\phi_a-\varpi(a)\|+\|\phi_b-\varpi(b)\|<1+1=2$ which implies $a=b$ because otherwise $\|\phi_a-\phi_b\|=2$.
Hence $S$ is uncountable.