= Solution
Let $\mathcal H$ be the family of <closed half-spaces> containing $C$. Certainly $C\subseteq\bigcap_{H\in\mathcal H}H$. To prove the reverse inclusion, exclude an arbitrary $z\notin C$.
Suppose first that $C$ is nonempty. Its <Euclidean projection onto a convex set> $p$ exists: minimizing the continuous squared distance may be restricted to a sufficiently large closed ball, whose intersection with the closed set is compact. Put $d=z-p\ne0$. For every $x\in C$, <convexity> puts $p+t(x-p)$ in $C$ for $0\leq t\leq1$. Minimality of $p$ gives
$$
0\leq\left.\frac d{dt}\|z-p-t(x-p)\|^2\right|_{t=0+}
=-2\langle d,x-p\rangle.
$$
Thus the <closed half-space>
$$
H_z=\{x:\langle d,x\rangle\leq\langle d,p\rangle\}
$$
contains $C$, while $\langle d,z\rangle=\langle d,p\rangle+\|d\|^2$ excludes $z$. Since every point outside $C$ is excluded by some member of $\mathcal H$,
$$
\boxed{C=\bigcap_{\substack{H\text{ a closed half-space}\\C\subseteq H}}H.}
$$
This is the <half-space representation of a closed convex set>. If $C=\varnothing$, every point can again be excluded by a containing half-space, so the intersection is empty. If $C=\mathbb R^n$, no proper half-space contains it and the empty intersection is, by convention, $\mathbb R^n$.
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