You might find enlightening the following sketched proof that $\, \mathbb Z[w],\ w = (1 + \sqrt{-19})/2\,$ is a non-Euclidean PID -- based on a sketch by the eminent number theorist Hendrik W. Lenstra.
Note that the proof in Dummit & Foote uses the Dedekind-Hasse criterion
to prove it is a PID, and the universal side divisor criterion
to prove it is not Euclidean is probably the simplest known.
The so-called universal side divisor criterion is essentially a
special case of research of Lenstra, Motzkin, Samuel, Williams
et al. that applies in much wider generality to Euclidean domains.
You can obtain a deeper understanding of Euclidean domains from
the excellent surveys by Lenstra in Mathematical Intelligencer
1979/1980 (Euclidean Number Fields 1,2,3) and Lemmermeyer's superb survey The Euclidean algorithm in algebraic number fields. Below is said sketched proof of Lenstra, excerpted from George Bergman's web page.
Let $\,w\,$ denote the complex number $\,(1 + \sqrt{-19})/2,\,$ and $\,R\,$ the ring $\, \Bbb Z[w].$
We shall show that $\,R\,$ is a principal ideal domain, but not a Euclidean
ring. This is Exercise III.3.8 of Hungerford's Algebra (2nd edition),
but no hints are given there; the proof outlined here was sketched for
me (Bergman) by H. W. Lenstra, Jr.
$(1)\ $ Verify that $\, w^2\! - w + 5 = 0,\,$ that $\,R = \{m + n\ a\ :\ m, n \in \mathbb Z\} = \{m + n\ \bar a\ :\ m, n \in \mathbb Z\},\,$ where the bar denotes complex conjugation, and that the map $\,x \to |x|^2 = x \bar x\,$ is nonnegative integer-valued
and respects multiplication.
$(2)\ $ Deduce that $\,|x|^2 = 1\,$ for all units of $\,R,\,$ and using a lower bound on the absolute value of the imaginary part of any nonreal member of $\,R,\,$ conclude that the only units of $\,R\,$ are $\pm 1.$
$(3)\ $ Assuming $\,R\,$ has a Euclidean function $\,h,\,$ let $\,x\ne 0\,$ be a nonunit of $\,R\,$ minimizing $\, h(x).\,$ Show that $\,R/xR\,$ consists of the images in this ring of $\,0\,$ and the units of $\,R,\,$ hence has cardinality at most $\,3.\,$ What nonzero rings are there of such cardinalities? Show $\,w^2 - w + 5 = 0 \,$ has no solution in any of these rings, and deduce a contradiction, showing that $\,R\,$ is not Euclidean.
We shall now show that $\,R\,$ is a principal ideal domain. To do this, let
$\,I\,$ be any nonzero ideal of $\,R,\,$ and $\,x\,$ a nonzero element of $\,I\,$ of least absolute value, i.e., minimizing the integer $\,x \bar x.\,$ We shall prove $\,I = xR.\,$ (Thus, we are using the function $\,x \to x \bar x\,$ as a substitute for a Euclidean function, even though it doesn't enjoy all such properties.)
For convenience, let us "normalize" our problem by taking $\,J = x^{-1}I.\,$ Thus, $\,J\,$ is an $\,R$-submodule of $\,\mathbb C,\,$ containing $\,R\,$ and having no nonzero element of absolute value $< 1.\,$ We shall show from these properties that $\, J - R = \emptyset,\,$ i.e. that $\,J = R.$
$(4)\ $ Show that any element of $\,J\,$ that has distance less than $\,1\,$ from some element of $\,R\,$ must belong to $\,R.\,$ Deduce that in any element of $\,J - R,\,$ the imaginary part must differ from any integral multiple of $\,\sqrt{19}/2\,$ by at least $\,\sqrt{3}/2.\,$ (Suggestion: draw a picture showing the set of complex numbers which the preceding observation excludes. However, unless you are told the contrary, this picture does not replace a proof; it is merely to help you find a proof.)
$(5)\ $ Deduce that if $\, J - R\,$ is nonempty, it must contain an element $\,y\,$ with imaginary part in the range $\,[\sqrt{3}/2,\,\sqrt{19}/2 - \sqrt{3}/2],\,$ and real part in the range $\, (-1/2,\,1/2].$
$(6)\ $ Show that for such a $\, y,\,$ the element $\, 2y\,$ will have imaginary part too close to $\,\sqrt{19}/2\,$ to lie in $\, J - R.\,$ Deduce that $\,y = w/2\,$ or $\,-\bar w/2,\,$ and hence that $\,w\,\bar w/2\,\in J.$
$(7)\ $ Compute $\, w\,\bar w/2,\,$ and obtain a contradiction. Conclude that $\,R\,$ is a principal ideal domain.
$(8)\ $ What goes wrong with these arguments if we replace $19$ throughout by $17$?
By $23$?
Here's what I think is a nice way to find a few PIDs that aren't Euclidean. It's not quite elementary, but if you know a bit about number fields I think it's a lot easier and nicer than the normal drudge.
Let $K=\mathbb{Q}(\sqrt{-d})$ for $d>3$ squarefree, with ring of integers $\mathcal{O}_K$. Then the only units in $\mathcal{O}_K$ are $\pm1$.
Suppose $\mathcal{O}_K$ is Euclidean with Euclidean function $\varphi$. Then take $x \in \mathcal{O}_K\setminus\{0,\pm1\}$ with $\varphi(x)$ minimal. By definition, any element of $\mathcal{O}_K$ can be written in the form $px+r$ where $\varphi(r) < \varphi(x)$, so it must be that $r \in \{0,\pm1\}$, i.e. $|\mathcal{O}_K/(x)|$ is $2$ or $3$. In other words $\mathcal{O}_K$ has a principal ideal of norm $2$ or $3$.
So now we know that if $K = \mathbb{Q}(\sqrt{-d})$ has class number one, where $d>3$ is squarefree*, $K$ is a non-Euclidean PID if there are no elements in $\mathcal{O}_K$ of norm $\pm2$ or $\pm3$. As $K$ is a PID (and degree $2$ over $\mathbb{Q}$), this is equivalent to saying that $2$ and $3$ are inert. To find some examples then:
If $d = 3\pmod{4}$, $\mathcal{O}_K = \mathbb{Z}[\frac{1+\sqrt{-d}}{2}]$, the minimal polynomial of $\frac{1+\sqrt{-d}}{2}$ over $\mathbb{Q}$ is $f_d(X)=X^2-X+\frac{1+d}{4}$. Applying Dedekind's criterion gives that $d$ works provided that $f_d(X)$ is irreducible $\pmod{2}$ and $\pmod{3}$. This then gives that $d = 19$ works (which is the usual example), but also shows that $d = 43,67$ or $163$ work as well (I think!).
*It can be shown that this implies $d \in \{1,2,3,7,11,19,43,67,163\}$
Best Answer
The method suggested by your professor is sometimes called the universal side divisor criterion for showing that a ring in non-Euclidean. The basic idea is quite simple. To prove that a domain is non-Euclidean it suffices to show that it lacks some property possessed by all Euclidean domain. Here this property is that if $x$ is a nonunit of minimal Euclidean value, then, modulo $x$, every element has remainder $0$ or a unit (else the remainder is a nonunit having Euclidean value smaller than $x$, contra hypothesis).
Based on your hint, in your case it seems that the only units of $R$ are $\pm 1$, so every element $w$ has remainder $\,0\,$ or $\,\pm1,\,$ i.e. $x$ divides $w$ or $w\mp1$. It remains to show that there is some $w$ for which this fails, thus completing the proof that $R$ is not Euclidean.
If you google "universal side divisor" you will find some insightful expositions, including worked examples. For example see Keith Conrad's Remarks about Euclidean domains. See also my answer here which outlines a proof that $\rm\: \mathbb Z[w],\ w = (1 + \sqrt{-19})/2\ $ is a non-Euclidean PID - based on sketch by Hendrik W. Lenstra.