This picture is an example in Dummit and Foote's Abstract Algebra. It shows that $\mathbb{Z}[\sqrt{-5}]$ is not a Euclidean domain by showing that the field norm $N(a+b\sqrt{-5})=a^2+5b^2$ doesn't allow the euclidean algorithm.
But why only showing the field norm not okay is enough?
On
Your phrase "by showing that the field norm ... doesn't allow the euclidean algorithm" is inaccurate.
Note that all Euclidean Domains are PIDs. Consequently, if $R = \Bbb{Z}(\sqrt{-5})$ is not a PID, it is not an ED.
To show that $R$ is not a PID, it is sufficient to exhibit a single ideal that is not principal. $I$ is a proposed ideal. The argument then uses the norm to constrain how $I$ could be generated by a single element of $R$. (It is convenient that number fields have a norm. This norm is inherited from $\Bbb{Q}(\sqrt{-5})$.) After eliminating every possible way $I$ could be principal, we conclude that $R$ is not an ED.
On
A little further ahead in Section 8.3 on UFDs there is another proof that avoids the field norm altogether and I think is probably a simpler argument depending on if you feel comfortable with maximal and prime ideals (proof of Proposition 11).
Given the definitions in D&F:
(1) Suppose $r\in R$ is nonzero and is not a unit. Then $r$ is called irreducible in $R$ if whenever $r=ab$ with $a,b \in R$, at least one of $a$ or $b$ must be a unit in $R$. Otherwise $r$ is said to be reducible.
(2) The nonzero element $p\in R$ is called prime in $R$ if the ideal $(p)$ generated by $p$ is a prime ideal. In other words, a nonzero element $p$ is a prime if it is not a unit and whenever $p|ab$ for any $a,b\in R$, then either $p|a$ or $p|b$.
and
Proposition 11. In a Principal Ideal Domain a nonzero element is a prime if and only if it is irreducible.
The calculations in the quoted section are actually useful in that they show that $3$ is irreducible, by using the field norm to show that if $3=\alpha\beta$ then $\alpha \lor \beta = \pm 1$.
The field norm can be used to show that the units of $\mathbb{Z}[\sqrt{-5}]$ are $\pm 1$, since the element $\alpha$ is a unit in $\mathcal{O}$ if and only if $N(\alpha)=\pm 1$, $a^2+5b^2=1$ ($-1$ is not possible because $a^2+5b^2\geq 0$), has only solutions $a=\pm 1$ because if $b\geq 1$ then $a^2+5b^2 \geq a^2 + 5 \geq 5$.
$N(3)=9=N(\alpha)N(\beta)=N(\alpha)N(a+b\sqrt{-5})=N(\alpha)(a^2+5b^2)$ and the calculations show that either $a^2+5b^2 = 9$ which would imply $\alpha = \pm 1$, or $a^2+5b^2 =\pm 1$, in either case $3$ is irreducible, while $3|(2+\sqrt{-5})(2-\sqrt{-5})=9$ but $3 \nmid (2+\sqrt{-5})$ and $3 \nmid (2-\sqrt{-5})$. Therefore $3$ is not prime.
$\mathbb{Z}[\sqrt{-5}]$ cannot be a PID because there exists an element which is irreducible and not prime, which contradicts Proposition 11.
Read the proof more carefully. They actually show that $I$ is not principal (using methods related to the field norm) and then use the fact that any euclidean domain is a principal ideal domain.