[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]
