[proofplan] We first show that every point of $K$ satisfies all inequalities defining the intersection, directly from the definition of the support function. For the reverse inclusion, we take a point $y \notin K$ and construct an explicit separating direction using a nearest point of $K$ to $y$. This produces a vector $u$ for which $\langle y,u\rangle>h_K(u)$, so $y$ is excluded. Finally, the positive homogeneity of $h_K$ shows that non-zero directions may be normalized to the unit sphere. [/proofplan]

[step:Show that points of $K$ satisfy every support inequality] Let \begin{align*} A:=\bigcap_{u\in \mathbb R^n}\{x\in \mathbb R^n: \langle x,u\rangle \le h_K(u)\}. \end{align*} Fix $x\in K$ and $u\in \mathbb R^n$. Since $h_K(u)=\sup_{z\in K}\langle z,u\rangle$ and $x\in K$, we have \begin{align*} \langle x,u\rangle \le \sup_{z\in K}\langle z,u\rangle=h_K(u). \end{align*} Thus $x\in A$. Since $x\in K$ was arbitrary, $K\subset A$. [/step]

[step:Construct a separating support direction for a point outside $K$] Fix $y\in \mathbb R^n\setminus K$. Define \begin{align*} \phi:K&\to \mathbb R\\ z&\mapsto |y-z|^2 . \end{align*} The function $\phi$ is continuous, and $K$ is compact and non-empty, so there exists $x_0\in K$ such that \begin{align*} |y-x_0|^2=\min_{z\in K}|y-z|^2. \end{align*} Since $y\notin K$, we have $y\ne x_0$. Define $u_0:=y-x_0\in \mathbb R^n\setminus\{0\}$. We claim that $\langle x-x_0,u_0\rangle\le 0$ for every $x\in K$. Fix $x\in K$ and define \begin{align*} \gamma:[0,1]&\to K\\ t&\mapsto (1-t)x_0+tx . \end{align*} This map has image in $K$ because $K$ is convex. By the minimizing property of $x_0$, \begin{align*} |y-x_0|^2\le |y-\gamma(t)|^2 \end{align*} for every $t\in[0,1]$. Since $y-\gamma(t)=u_0-t(x-x_0)$, this gives \begin{align*} |u_0|^2\le |u_0-t(x-x_0)|^2 =|u_0|^2-2t\langle u_0,x-x_0\rangle+t^2|x-x_0|^2 . \end{align*} Hence, for every $t\in(0,1]$, \begin{align*} 2\langle u_0,x-x_0\rangle\le t|x-x_0|^2. \end{align*} Letting $t\to 0^+$ yields \begin{align*} \langle x-x_0,u_0\rangle\le 0. \end{align*} Therefore, for every $x\in K$, \begin{align*} \langle x,u_0\rangle\le \langle x_0,u_0\rangle. \end{align*} Taking the supremum over $x\in K$ gives \begin{align*} h_K(u_0)\le \langle x_0,u_0\rangle. \end{align*} On the other hand, \begin{align*} \langle y,u_0\rangle =\langle x_0+u_0,u_0\rangle =\langle x_0,u_0\rangle+|u_0|^2 >\langle x_0,u_0\rangle . \end{align*} Thus \begin{align*} h_K(u_0)<\langle y,u_0\rangle. \end{align*} So $y\notin A$. [/step]

[step:Conclude equality for the intersection over all directions] The previous step shows that every $y\in\mathbb R^n\setminus K$ fails at least one inequality defining $A$. Therefore $A\subset K$. Combined with $K\subset A$, this proves \begin{align*} K=\bigcap_{u\in \mathbb R^n}\{x\in \mathbb R^n: \langle x,u\rangle \le h_K(u)\}. \end{align*} [/step]

[step:Normalize non-zero directions to obtain the sphere intersection] Let \begin{align*} B:=\bigcap_{v\in S^{n-1}}\{x\in \mathbb R^n: \langle x,v\rangle \le h_K(v)\}. \end{align*} Since $S^{n-1}\subset \mathbb R^n$, the intersection over all of $\mathbb R^n$ is contained in the intersection over $S^{n-1}$, so $K=A\subset B$. It remains to prove $B\subset K$. Let $y\in\mathbb R^n\setminus K$. From the previous separation construction, there exists $u_0\in\mathbb R^n\setminus\{0\}$ such that \begin{align*} h_K(u_0)<\langle y,u_0\rangle. \end{align*} Define $v_0:=u_0/|u_0|\in S^{n-1}$. Since $u_0=|u_0|v_0$ and $|u_0|>0$, the support function is positively homogeneous in this direction: \begin{align*} h_K(u_0) =\sup_{z\in K}\langle z,|u_0|v_0\rangle =|u_0|\sup_{z\in K}\langle z,v_0\rangle =|u_0|h_K(v_0). \end{align*} Also \begin{align*} \langle y,u_0\rangle=|u_0|\langle y,v_0\rangle. \end{align*} Dividing the strict inequality $h_K(u_0)<\langle y,u_0\rangle$ by $|u_0|>0$ gives \begin{align*} h_K(v_0)<\langle y,v_0\rangle. \end{align*} Thus $y\notin B$. Since every $y\notin K$ is excluded from $B$, we have $B\subset K$. Therefore \begin{align*} K=\bigcap_{v\in S^{n-1}}\{x\in \mathbb R^n: \langle x,v\rangle \le h_K(v)\}. \end{align*} [/step]