[proofplan]
We prove the two inclusions separately. The inclusion $K$ into the intersection follows from the definition of $h_K$ as the supremum of the linear functional $z \mapsto u \cdot z$ over $K$. For the reverse inclusion, we take a point $y \notin K$, choose a nearest point $p \in K$ to $y$, and use convexity of $K$ to show that the vector $u := y - p$ strictly separates $y$ from $K$. This produces one half-space in the displayed intersection that excludes $y$.
[/proofplan]
[step:Verify that every point of $K$ satisfies every support inequality]
Let
\begin{align*}
S := \bigcap_{u \in \mathbb{R}^n} \{x \in \mathbb{R}^n : u \cdot x \le h_K(u)\}.
\end{align*}
We first prove $K \subset S$.
Fix $x \in K$ and $u \in \mathbb{R}^n$. Since $h_K(u)$ is the supremum of the set $\{u \cdot z : z \in K\} \subset \mathbb{R}$ and since $x \in K$, we have
\begin{align*}
u \cdot x \le \sup_{z \in K} u \cdot z = h_K(u).
\end{align*}
Because this holds for every $u \in \mathbb{R}^n$, the point $x$ belongs to every half-space in the intersection defining $S$. Hence $x \in S$. Since $x \in K$ was arbitrary, $K \subset S$.
[/step]
[step:Choose a nearest point of $K$ to a point outside $K$]
We prove $S \subset K$ by showing that every point outside $K$ is outside $S$.
Let $y \in \mathbb{R}^n \setminus K$. Define
\begin{align*}
d := \inf_{z \in K} |y - z|.
\end{align*}
Because $K$ is compact and the function
\begin{align*}
\psi: K &\to \mathbb{R} \\
z &\mapsto |y - z|
\end{align*}
is continuous, there exists $p \in K$ such that
\begin{align*}
|y - p| = d.
\end{align*}
Since $K$ is compact, it is closed in $\mathbb{R}^n$. Therefore $y \notin K$ implies $d > 0$, and hence $y \ne p$.
Define the vector
\begin{align*}
u := y - p \in \mathbb{R}^n.
\end{align*}
Then $u \ne 0$ and
\begin{align*}
u \cdot y - u \cdot p
= (y - p) \cdot y - (y - p) \cdot p
= (y - p) \cdot (y - p)
= |y - p|^2
> 0.
\end{align*}
Thus
\begin{align*}
u \cdot p < u \cdot y.
\end{align*}
[/step]
[step:Use convexity to show that the nearest point defines a supporting half-space]
We claim that
\begin{align*}
u \cdot x \le u \cdot p
\end{align*}
for every $x \in K$.
Fix $x \in K$. Since $K$ is convex and $p \in K$, the line segment from $p$ to $x$ lies in $K$. Thus for every $t \in [0,1]$,
\begin{align*}
p + t(x - p) \in K.
\end{align*}
Define
\begin{align*}
\varphi: [0,1] &\to \mathbb{R} \\
t &\mapsto |y - p - t(x - p)|^2.
\end{align*}
Because $p$ minimizes the distance from $y$ to points of $K$, and because $p + t(x-p) \in K$, we have
\begin{align*}
\varphi(0) = |y - p|^2 \le |y - p - t(x - p)|^2 = \varphi(t)
\end{align*}
for every $t \in [0,1]$.
Expanding the square gives
\begin{align*}
\varphi(t)
&= |y - p|^2 - 2t (y - p) \cdot (x - p) + t^2 |x - p|^2 \\
&= \varphi(0) - 2t u \cdot (x - p) + t^2 |x - p|^2.
\end{align*}
Therefore, for every $t \in (0,1]$,
\begin{align*}
0 \le \frac{\varphi(t) - \varphi(0)}{t}
= -2u \cdot (x - p) + t |x - p|^2.
\end{align*}
Taking the limit as $t \to 0^+$ yields
\begin{align*}
0 \le -2u \cdot (x - p).
\end{align*}
Hence
\begin{align*}
u \cdot (x - p) \le 0,
\end{align*}
which is equivalent to
\begin{align*}
u \cdot x \le u \cdot p.
\end{align*}
Since $x \in K$ was arbitrary, the inequality holds for every $x \in K$.
[/step]
[step:Exclude the outside point from the intersection]
From the preceding step,
\begin{align*}
u \cdot x \le u \cdot p
\end{align*}
for every $x \in K$. Taking the supremum over $x \in K$ gives
\begin{align*}
h_K(u) = \sup_{x \in K} u \cdot x \le u \cdot p.
\end{align*}
From the construction of $u$, we also have
\begin{align*}
u \cdot p < u \cdot y.
\end{align*}
Combining these inequalities,
\begin{align*}
h_K(u) < u \cdot y.
\end{align*}
Thus $y$ does not satisfy the half-space inequality corresponding to this vector $u$, namely
\begin{align*}
u \cdot y \le h_K(u).
\end{align*}
Therefore $y \notin S$.
We have shown that every $y \in \mathbb{R}^n \setminus K$ lies outside $S$, so $S \subset K$. Together with $K \subset S$, this proves
\begin{align*}
K = \bigcap_{u \in \mathbb{R}^n} \{x \in \mathbb{R}^n : u \cdot x \le h_K(u)\}.
\end{align*}
[/step]
