Diagonalization gives a map $f\colon U(n)\to \mathbb{T}^n/S_n$. The map $f$ is also the projection to the orbit-space of of the $U(n)$-action by conjugacy on $U(n)$. Hence $f$ is a submetry; i.e., $f(B_r(M))=B_r(f(M))$ for any matrix $M$. Hence your statement follows.

Diagonalization gives a map $f\colon U(n)\to \mathbb{T}^n/S_n$. The map $f$ is also the projection to the orbit-space of the $U(n)$-action by conjugacy on $U(n)$. Hence $f$ is a submetry; i.e., $f(B_r(M))=B_r(f(M))$ for any matrix $M$. Hence your statement follows.