If $R$ is not a field, then it has nonunits. Consider the set $S=\{\varphi(a)\mid a\text{ is not a unit, and }a\neq 0\}$, where $\varphi$ is the Euclidean function.
It is a nonempty set of positive integers. By the Least Element Principle, it has a smallest element. Let $c\in R$ be a nonunit, nonzero, such that $\varphi(c)$ is the smallest element of $S$.
Edited. I claim that $c$ satisfies the conditions of the problem. Let $a\in R$. Then we can write $a = qc + r$, with either $r=0$ or $\varphi(r)\lt \varphi(c)$. If $r=0$, we are done. If $r\neq 0$, then $\varphi(r)\lt \varphi(c)$, then $\varphi(r)\notin S$, hence $r$ does not satisfy the condition
$r$ is not a unit, and $r\neq 0$.
Since $r\neq 0$, it follows that $r$ must be a unit.
Thus, for every $a\in R$, there exist $q,r\in R$ such that $a=qc+r$, and either $r=0$ or $r$ is a unit, as desired.
If $\mathbb{Z}[\sqrt{-10}]$ were an Euclidean domain, then it would be a UFD.
By considering norms, we see that $2$, $5$, and $\sqrt{-10}$ are irreducible in $\mathbb{Z}[\sqrt{-10}]$.
Since $10 = (-1)(\sqrt{-10})^2 = 2 \times 5$ are two distinct factorizations into irreducibles, $\mathbb{Z}[\sqrt{-10}]$ is not a UFD and so cannot be an Euclidean domain.
Best Answer
Note that if $u =a+b\sqrt{2} \in \Bbb{Z}[\sqrt{2}]$ then the norm of $u$ is defined as $$v(u)=\vert \vert u \vert \vert=\vert a^2-2b^2 \vert$$
The half of the battle is already done by you. For the other part, note that $1=v(1)$ is always the minimum ,since $$v(1) \leq v(1 \cdot u) =v(u)$$