Let $x_0 \in \mathbb R^n$, and let $\mathcal K$ denote the set of compact convex subsets of $\mathbb R^n$ equipped with the Hausdorff metric. Consider the map $f: K \mapsto \Pi_K x_0$, where $\Pi_K$ denotes orthogonal projection onto $K$. It is not the case that $f$ is Lipschitz. To see this, consider the case where $n = 2$, $x_0 = 0$, $K$ is the line segment $[(R, -1), (R, 1)]$, and $K'$ is the line segment $[(R-t, -1), (R+t, 1)]$, where $t = \frac{R}{2}(1 - \sqrt{1 - 4 / R^2}) \sim 1/R$. Then $d_H(K, K') = t$, but $|\Pi_K x_0 - \Pi_{K'} x_0| \geq 1$.
For $\epsilon > 0$, suppose $K, K' \subset B(0, 1)$, $x_0 = 0$, and $\min(\text{diam}(K), \text{diam}(K')) \geq \epsilon$. Does there exist a constant $C$ depending on $\epsilon$ and $n$ only such that $|f(K) - f(K')| \leq C d_H(K, K')$? How does $C$ behave as $\epsilon \to 0$? The example above shows that it must be at least $\sim 1 / \epsilon$.
