Proof of Strong Separation of a Point from a Closed Convex Subset of $\mathbb{R}^n$

We first prove (i), distinguishing whether $C$ is empty. If $C=\emptyset$, take for $a$ the first standard basis vector $e_{1}$, which is nonzero since its first coordinate is $1$; the inequality on $C$ then holds vacuously. If $C\neq\emptyset$, let $p\in C$ be the point of $C$ nearest to $x$, given by Nearest Points in a Nonempty Closed Convex Subset of $\mathbb{R}^n$: Existence, Uniqueness and the Variational Inequality (i), and put $a=x-p$; then $a\neq0$ because $p\in C$ and $x\notin C$. By Nearest Points in a Nonempty Closed Convex Subset of $\mathbb{R}^n$: Existence, Uniqueness and the Variational Inequality (ii), $(x-p)\cdot(y-p)\le0$ for every $y\in C$; that is, $a\cdot y\le a\cdot p$, and by bilinearity

$$a\cdot y\le a\cdot p=a\cdot x-a\cdot(x-p)=a\cdot x-\lVert a\rVert^{2}\qquad\text{for every }y\in C.$$

In both cases $\lVert a\rVert^{2}>0$ because $a\neq0$, so $a\cdot x-\lVert a\rVert^{2}<a\cdot x$, which completes the proof of (i).

For (ii), let $a$ be as in (i), and put

$$\alpha=a\cdot x-\tfrac{1}{2}\lVert a\rVert^{2},\qquad\varepsilon=\tfrac{1}{2}\lVert a\rVert^{2}>0.$$

Then

$$a\cdot y\le a\cdot x-\lVert a\rVert^{2}=\alpha-\varepsilon\quad\text{for every }y\in C,\qquad a\cdot x=\alpha+\varepsilon.$$

By Separating Hyperplanes in $\mathbb{R}^n$: Separation, Strict Separation and Strong Separation (iii), this says that $H(a,\alpha)$ strongly separates $C$ and $\{x\}$.