We have the norm (even though it is not always positive) obtained by multiplying $a+b\sqrt {10}$ with its conjugate $a-b\sqrt {10}$, that is $N(a+b\sqrt{10})=a^2-10b^2\in\mathbb Z$. So we have $N(9)=81$ and $N(2+2\sqrt{10})=-36$. We can restrict our search to elements of norm dividing both $81$ and $36$, that is we must have $N(a+b\sqrt{10})\in\{\pm1,\pm3,\pm9\}$. Looking for small solutions of this you will stumble upon $1+\sqrt {10}$, which is both a divisor of $9$ and $2+2\sqrt{10}$ and a linear combination of these, namely $(1+\sqrt {10})\cdot 9-4\cdot (2+2\sqrt{10})$.
The structure of the proof is to show that $\mathbb{Z}[\sqrt{-5}]$ is not a PID, and therefore not a Euclidean Domain. To show that it is not a PID, it suffices to find an ideal that is not principal. The ideal $I = (3,2 +\sqrt{-5})$ is shown not to be principal by showing that $3$ and $2+\sqrt{-5}$ cannot be viewed as multiples (in the ring) of fixed element $a + b\sqrt{-5}$, and this is ruled out by using a norm argument. A different norm might have worked, what's important is that the norm has a nice symmetry with respect to multiplication, and this reduces the original problem to a problem with ordinary integers which is more easily shown to be unsolvable.
To take a simpler example, with vector spaces: suppose you have three vectors $u,v,w$ and you want to show $u \neq v+w$. It suffices to find at least one linear transformation for which $T(u) \neq T(v) + T(w)$, and this might be an easier than inspecting the original vectors if the vector space is very complicated.
An even simpler example: suppose you have two functions $f$ and $g$, and you want to show $f$ is not not a linear multiple of $g$, i.e. $f \neq cg$. What you can do is take two points $x,y$ and consider the system of equations:
$$ f(x) = cg(x)$$
$$ f(y) = cg(y)$$
If this system has no solution, you have shown $f \neq cg$. You might complain "hey! what if you tried other points? maybe $f = cg$ with respect to different points!" but (1) $f = cg$ is a general statement, it is not "with respect to particular points" and (2) finding one counterexample to $f=cg$ suffices.
Best Answer
One such simple example of a non-Euclidean PID is $ K[[x,y]][1/(x^2\!+\!y^3)]\,$ over any field $\,K,\,$ i.e. adjoin the inverse of $\,x^2\!+\!y^3$ to a bivariate power series ring over a field. For the proof, and a general construction method see
D.D. Anderson. An existence theorem for non-euclidean PID’s,
Communications in Algebra, 16:6, 1221-1229, 1988.
For number rings, by Weinberger (1973), assuming GRH, a UFD number ring R with infinitely many units is Euclidean, e.g. real quadratic number rings are Euclidean $\!\iff\!$ PID $\!\iff\!$ UFD.