Perhaps my notation is confusing or there is an error in terminology so let me clarify the problem first. I have some convex set X embedded inside a larger set A. So X \subset A. I have elements \rho \in A. The function D(\rho, \sigma) = ||\rho - \sigma|| finds \sigma \in X such that the distance (any valid metric) from \rho to \sigma is minimized.
It is the shortest line from \rho to X.
Proof that this is convex:
Consider \rho_1, \rho_2 \in A and let \sigma_1, \sigma_2 \in X be the respective closest points.
For any \lambda\rho_1 + (1-\lambda)\rho_2, we want to show that there exists \sigma_3\in X such that ||\lambda\rho_1 + (1-\lambda)\rho_2 - \sigma_3|| \leq \lambda ||\rho_1 - \sigma_1|| + (1-\lambda)||\rho_2 - \sigma_2||
Let \sigma_3 = \lambda \sigma_1 + (1-\lambda)\sigma_2. Since X is convex, \sigma_3\in X. Then we have,
||\lambda\rho_1 + (1-\lambda)\rho_2 - \sigma_3|| = ||\lambda(\rho_1 - \sigma_1) + (1-\lambda)(\rho_2 - \sigma_2)|| \leq \lambda ||\rho_1 - \sigma_1|| + (1-\lambda)||\rho_2 - \sigma_2||.
The last line uses a triangle inequality.