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I would say that $D_\mu '\psi' = U(D_\mu \psi)$ is a requirement, rather than an assumption: When $U$ is a global transformation (i.e., when $\lambda$ is independent of $x$), we have trivially $\partial_\mu \psi' = U(\partial_\mu \psi)$, and so $(\partial_\mu \psi)^\dagger (\partial_\mu \psi) = (\partial_\mu \psi')^\dagger (\partial_\mu \psi')$. We define ...