$X$ is a normed space and $C\subset X$ a convex and closed cone ($\lambda C\subset C\ \forall \lambda\geq0$), and $C':=\{x'\in X':x'(x)\geq0\ \forall x\in C\}$ the dual cone of $C$.
I want to show:
$(i):\quad C\neq X\Rightarrow C'\neq\{0\}$
$(ii):\quad x'(x)\geq0\ \forall x'\in C'\Rightarrow x\in C$
Because in class we talked about separation theorems, my approach so far was:
Let $C\neq X$ then $\exists x_0\notin C$. Then $\{x_0\}$ is closed and convex, the Hahn-Banach-Separation-Theorem provides $x_0'$ such that $x_0'(x_0)<\inf_{x\in C}x_0'(x)\leq0$.
This is were I'm stuck...
Your approach is good, you just need to use the cone property of $C$: By the separation theorem there exist $a\in \mathbb{R}$ and $x_0^*\in X^*$ such that $$x_0^*(x_0)<a<x_0^*(x), \ \forall x\in C.$$
If there existed an $y_0\in C$ such that $x_0^*(y_0)<0$, then $ny_0\in C$ for every $n\in \mathbb{N}$, so $\lim_{n\rightarrow \infty} x_0^*(ny_0) =-\infty$ which would contradict the fact that $x_0^*(x_0)<a<x_0^*(x), \ \forall x\in C.$ So $x_0^*(x)\geq 0$, for every $x\in C$. Since $x_0^*(0)=0$ and $0\in C$, we have that $a<0$. This solves both (i) and (ii) simultaneously.