Let $C \subseteq \mathbb{R}^n$ be a closed convex set, and $x^* \in C^c$ (not in $C$ and its closure).
Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=\min_{z \in C}\|z -x^*\|_2$.
Consider the closed half-space $H:=\{x \in \mathbb{R}^n: \langle a,x\rangle \leq \gamma \}$, where $a=x^*-\prod_C(x^*)$ and $\gamma=\langle a,\prod_C(x^*)\rangle$ where $\prod_C(x^*)$ is projection of $x^*$ onto $C$.
Show that $d_H(x^*)=d_C(x^*)$.
Intuitively, it says that the hyperplane defining in this way is tangent to the set $C$ at point $\prod_C(x^*)$.
Extension:
Show that the minimum distance from $x^*$ to $C$ is the maximum among the distance from $x^*$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^*)=\max_{C \subseteq H}d_H(x^*)$ where the max is taken over all closed half-spaces $H$'s containing $C$.
Best Answer
Note that $\Pi_C(x^\ast) \in C\subseteq H$, and $\Pi_H(x^\ast) \in H$. Then by construction of $H$, $$0 \geq \langle x^\ast - \Pi_C(x^\ast),\Pi_H(x^\ast) - \Pi_C(x^\ast)\rangle = \langle \Pi_C(x^\ast) - x^\ast, \Pi_C(x^\ast) - \Pi_H(x^\ast) \rangle,$$ and by the variational inequality of Euclidean distance $$0 \geq \langle x^\ast - \Pi_H(x^\ast),\Pi_C(x^\ast) - \Pi_H(x^\ast)\rangle.$$ So add the two inequalities together, and you should see that, in fact, $\Pi_C(x^\ast) = \Pi_H(x^\ast)$!