Let $X$ be a locally convex topological vector space and $C$ be a closed convex subset of $X.$ Then $C$ is the intersection of closed half-spaces, where the closed half-spaces are of the form $\left \{\xi\ |\ \mathfrak R\ \rho (\xi) \leq t \right \}$ for some $\rho \in X^{\ast}$ and $t \in \mathbb R.$
Let $H$ be the collection of all the closed half-spaces containing $C.$ Then clearly $C \subseteq H.$ If we can show that $C = H$ then we are done. If not, let us take $z \in H \setminus C.$ Then consider the sets $\{z\}$ and $C.$ The set $\{z\}$ is compact convex and $C$ is given to be closed convex and also they are disjoint. So by Hahn-Banach separation theorem there exist $\rho \in X^{\ast}$ and $t_1, t_2 \in \mathbb R$ such that $$\mathfrak R\ \rho (z) \lt t_1 \lt t_2 \lt \mathfrak R\ \rho (x)$$ for all $x \in C.$ I got stuck here and couldn't able to proceed further. Could anybody please give me some hint?
Thanks for your time.
You wrote: Let $H$ be the collection of closed half-spaces containing $C.$ I think you meant: Let $H$ be the intersection all closed half-spaces containing $C.$
Also you wrote $z \in Z\setminus C$ for $z \in H\setminus C$.
Consider $\{\xi: \Re \rho(\xi) \geq t_1\}$. This is a half-space containing $C$. [Note that $\{\xi: \Re \rho (\xi) \geq t_1\}$ is same as $\{\xi: \Re \rho' (\xi) \leq -t_1\}$ where $\rho'=-\rho$]. Hence, $H \subseteq \{\xi: \Re \rho(\xi) \geq t_1\}$. This is a contradiction because $z \in H$ and $\Re \rho (z) <t_1$.