Proof of A Closed Convex Subset of $\mathbb{R}^n$ Is the Intersection of the Closed Half-Spaces Containing It

Throughout, closedness of $C$ refers to the metric topology $\mathcal{T}_{d_{E}}$ of $\mathbb{R}^{n}$, which exists by The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula (ii), and a point $y$ lies in $H^{\le}(a,\alpha)$ exactly when $a\cdot y\le\alpha$, by Hyperplanes and Half-Spaces in $\mathbb{R}^n$ (ii). Let $E$ denote the set on the right-hand side of (i); we show that a point $x\in\mathbb{R}^{n}$ lies in $E$ exactly when it lies in $C$.

Let $x\in C$. For every $a\neq0$ and $\alpha$ with $C\subseteq H^{\le}(a,\alpha)$ we have $x\in H^{\le}(a,\alpha)$, that is, $a\cdot x\le\alpha$. So $x\in E$.

Let $x\notin C$. By Strong Separation of a Point from a Closed Convex Subset of $\mathbb{R}^n$ (i), applied to $C$ and $x$, there is $a\in\mathbb{R}^{n}$ with $a\neq0$ such that, with $\alpha=a\cdot x-\lVert a\rVert^{2}$,

$$a\cdot y\le\alpha\quad\text{for every }y\in C\qquad\text{and}\qquad a\cdot x>\alpha.$$

So $C\subseteq H^{\le}(a,\alpha)$ while $a\cdot x\le\alpha$ fails, and $x\notin E$. This proves (i).