 Solution

If $C={\mathbb{R}}^n$, the family of proper containing half-spaces is empty and its intersection is, by convention, ${\mathbb{R}}^n$. If $C=\varnothing$, every closed half-space contains it and their intersection is empty. Now suppose $C$ is a nonempty proper closed convex set. Take $x\notin C$ and let $z$ be its Euclidean projection onto a convex set. This projection exists: a minimizing sequence can be restricted to a bounded ball, and closedness gives attainment. It is unique by convexity and strict convexity of squared distance.

For $y\in C$, the segment $z+t(y-z)$ remains in $C$ for $0\le t\le 1$. Minimality at $t=0$ impliesThus the closed half-space $H_x=\{y:⟨x-z,y-z⟩\le 0\}$ contains $C$ but excludes $x$, since $⟨x-z,x-z⟩>0$. Every point outside $C$ is excluded by at least one containing half-space. The reverse inclusion is immediate, givingThis is the half-space representation of a closed convex set. The argument gives an explicit separating hyperplane rather than just citing a Hahn-Banach separation theorem.