Proof of The Separating Hyperplane Theorem in $\mathbb{R}^n$

Let $M=C-D=\{x-y:x\in C,\ y\in D\}$ be the difference set; it is convex by Operations Preserving Convexity in $\mathbb{R}^n$: Intersections, Translates, Scalar Multiples, Sums, Affine Images and Preimages, Closure and Interior (v). Since $C\cap D=\emptyset$, the origin does not lie in $M$, and since $\operatorname{int}M\subseteq M$ (Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (i)), it does not lie in $\operatorname{int}M$ either.

By The Supporting Hyperplane Theorem in $\mathbb{R}^n$ (i), applied to $M$ and the origin, there is $a\in\mathbb{S}$ with $a\cdot w\le a\cdot0=0$ for every $w\in M$. Taking $w=x-y$ and using bilinearity, we obtain

$$a\cdot x\le a\cdot y\qquad\text{for all }x\in C\text{ and }y\in D,$$

and $a\neq0$ because $\lVert a\rVert=1$.

It remains to choose the level $\alpha$. As $D$ is nonempty, fix $y_{0}\in D$. The set $T=\{a\cdot x:x\in C\}$ is nonempty, since $C$ is, and bounded above by $a\cdot y_{0}$, so it has a supremum $\alpha$ (The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vii)). As an upper bound of $T$, $\alpha\ge a\cdot x$ for every $x\in C$. For $y\in D$ the number $a\cdot y$ is an upper bound of $T$ by the inequality just displayed, so $\alpha\le a\cdot y$, the supremum being the least upper bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)). Hence

$$a\cdot x\le\alpha\le a\cdot y\qquad\text{for all }x\in C\text{ and }y\in D,$$

which is (i). Thus $C\subseteq H^{\le}(a,\alpha)$ and $D\subseteq H^{\ge}(a,\alpha)$ (Hyperplanes and Half-Spaces in $\mathbb{R}^n$ (ii)), and $H(a,\alpha)$ separates $C$ and $D$ by Separating Hyperplanes in $\mathbb{R}^n$: Separation, Strict Separation and Strong Separation (i).