To show $\mathbb{Z}[\sqrt{-5}]$ is not a Euclidean domain, why suffices to show only the field norm $N(a+b\sqrt{-5})=a^2+5b^2$ doesn't work? 
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? 
 A: 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.
A: 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.
A: 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.
