Overhead breaks an unrestricted diagonalization claim

Choose a valid coded machine $M$ that makes one transition and then halts, and give the fixed simulator $U_1$ the pair $⟨M,ϵ⟩$. The concrete code $<mi\; mathvariant="normal">⌜</mi>M<mi\; mathvariant="normal">⌝</mi>$ has a unary arity header followed by all self-delimiting description fields, so its complete decoding requires reading more than one input cell.

By definition, $U_1$ completes that decoding before it simulates the one transition of $M$. Merely reaching and reading the later code cells takes more than one transition on the simulator's input tape. Thus this valid one-step computation costs more than one simulator step, contradicting the claimed universal bound at $n=1$.