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

Throughout, closedness and compactness in $\mathbb{R}^{n}$ refer to the metric topology $\mathcal{T}_{d_{E}}$, which exists by The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula (ii), and for $b\in\mathbb{R}^{n}$ and $\beta\in\mathbb{R}$ a point $y$ lies in $H(b,\beta)$ exactly when $b\cdot y=\beta$, by Hyperplanes and Half-Spaces in $\mathbb{R}^n$ (i), and in $H^{\le}(b,\beta)$ exactly when $b\cdot y\le\beta$, by Hyperplanes and Half-Spaces in $\mathbb{R}^n$ (ii).

If $C=\emptyset$, every point of $\mathbb{S}$ satisfies (i), the inequality holding vacuously; for instance $e_{1}$, whose first coordinate is $1$ and whose other coordinates are $0$, so that $\lVert e_{1}\rVert^{2}=1$. Assume from now on that $C\neq\emptyset$.

For $z\in C$ let

$$A_{z}=\{a\in\mathbb{R}^{n}:(z-x)\cdot a\le0\},$$

which is closed by Sets Cut Out by Linear Inequalities in $\mathbb{R}^n$ Are Closed (i); by symmetry and bilinearity, $a\in A_{z}$ exactly when $a\cdot z\le a\cdot x$. So (i) asks for a point of $\mathbb{S}$ lying in every $A_{z}$, and by compactness of $\mathbb{S}$ it suffices to treat finitely many $z$ at a time.

Let $J\subseteq C$ be nonempty and finite (listed as $z_{1},\dots,z_{N}$ by a bijection $[N]\to J$, Finite Sets (i)), and let $Q=\operatorname{conv}J$. The set $Q$ is compact by The Convex Hull of a Compact Subset of $\mathbb{R}^n$ is Compact (ii), hence closed by Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (ii); it is convex with $J\subseteq Q\subseteq C$ by Convex Hull of a Subset of $\mathbb{R}^n$ (i), since $C$ is a convex superset of $J$. Since $Q\subseteq C$, also $\operatorname{int}Q\subseteq\operatorname{int}C$ (Interior of a Subset of a Metric Space (ii)), hence $x\notin\operatorname{int}Q$. By A Point Outside the Interior of a Closed Convex Subset of $\mathbb{R}^n$ Has a Supporting Unit Normal (i), applied to $Q$ and $x$, there is $a\in\mathbb{S}$ with $a\cdot z\le a\cdot x$ for every $z\in Q$, in particular for every $z\in J$. Hence

$$\mathbb{S}\cap\bigcap_{z\in J}A_{z}\neq\emptyset.$$

Now The Finite Intersection Property of a Compact Set (i), in $(\mathbb{R}^{n},\mathcal{T}_{d_{E}})$, applied to the compact set $\mathbb{S}$, the nonempty index set $C$ and the family $(A_{z})_{z\in C}$ of closed sets, whose hypothesis is the last display, gives $a\in\mathbb{S}$ with $a\in A_{z}$, that is $a\cdot z\le a\cdot x$, for every $z\in C$. This proves (i).

For (ii) and (iii), let $a$ be as in (i); then $a\neq0$ because $\lVert a\rVert=1$. Put $\alpha=a\cdot x$, so that $x\in H(a,\alpha)$ and $C\subseteq H^{\le}(a,\alpha)$. If $x\in C$, it follows that $H(a,\alpha)$ supports $C$ at $x$ by Supporting Hyperplane of a Subset of $\mathbb{R}^n$ at a Point (i), which is (ii). Next, every closure point of $C$ is a closure point of the larger set $H^{\le}(a,\alpha)$ (Closure of a Subset of a Metric Space (i)), and so lies in $H^{\le}(a,\alpha)$, this half-space being closed (Sets Cut Out by Linear Inequalities in $\mathbb{R}^n$ Are Closed (iii), Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (iv)). Hence $\operatorname{cl}C\subseteq H^{\le}(a,\alpha)$ (Closure of a Subset of a Metric Space (ii)), and if $x\in\operatorname{cl}C$, then $H(a,\alpha)$ supports $\operatorname{cl}C$ at $x$, which is (iii).