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$X$ 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...

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    Your progress on $(i)$ is incorrect, since a cone doesn't have to be convex, thus application of the separation theorem invalid. Consider (geometrically) $X=\mathbb R^2$, with $C=\{(x,0):x\geq0\}\cup\{(0,y):y\geq0\}$, and $x_0=(1,1)$. – 2017-01-16
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    You are correct. I forgot the assumption that the cone is also convex and closed. – 2017-01-16
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    I haven't gotten around to working on this, but try assuming $C'=\{0\}$ and showing that $C=X$. I think that might be the easiest way to proceed. – 2017-01-16

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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)

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)