Ring of rational integers is not uniquely Euclidean for any norm

Statement

The ring of rational integers cannot be made a uniquely Euclidean domain with any norm.

Proof

The proof can be broken down into the following steps:

1. We cannot have $a\neq 0$ such that $N(a)<N(1)$: If such an $a$ exists, we have $1=1.(1-a)+a$ and also $1=1.1+0$: two Euclidean divisions for $1$ by itself.
2. We cannot have $a\neq 0$ such that $N(a)<N(-1)$: If such an $a$ exists, we have $-1=-1.(1+a)+a$ and also $-1=-1.1+0$: two Euclidean divisions for $-1$ by itself.
3. Thus, $N(1)=N(-1)$ and no element has smaller norm.
4. Suppose $a\neq 0,-1$. Then at most one of these holds: $N(a)<N(a+1)$ or $N(-1)<N(a+1)$. This follows from the uniqueness of Euclidean division for $a$ by $a+1$.
5. $N(a)=N(1)$ for all positive integers $a$: We prove this by induction on $a$. Clearly, $N(1)=N(1)$. Suppose $N(a)=N(1)$.By the previous step, either $N(a)<N(a+1)$ or $N(-1)<N(a+1)$. But by the induction hypothesis, $N(a)=N(1)$, which in turn is equal to $N(-1)$, so $N(a)<N(a+1)\iff N(-1)<N(a+1)$. Thus, neither of these holds, so $N(a+1)\leq N(a)=N(1)$. Step (3) then forces $N(a+1)=N(1)$.
6. $N(-b)=N(1)$ for all positive integers $b$: The proof is analogous to the preceding one, relying on the division of $-b$ by $-b-1$.
7. Combining the last two steps, all elements have equal norm, and this is clearly not a Euclidean norm.