proof that a Euclidean domain is a PID

Let $D$ be a Euclidean domain, and let $𝔞\subseteq D$ be a nonzero ideal. We show that $𝔞$ is principal. Let

$$A=\{\nu (x):x\in 𝔞,x\ne 0\}$$

be the set of Euclidean valuations of the non-zero elements of $𝔞$. Since $A$ is a non-empty set of non-negative integers, it has a minimum $m$. Choose $d\in 𝔞$ such that $\nu (d)=m$. Claim that $𝔞=(d)$. Clearly $(d)\subseteq 𝔞$. To see the reverse inclusion, choose $x\in 𝔞$. Since $D$ is a Euclidean domain, there exist elements $y,r\in D$ such that

$$x=yd+r$$

with or $r=0$. Since $r\in 𝔞$ and $\nu (d)$ is minimal in $A$, we must have $r=0$. Thus $d|x$ and $x\in (d)$.