Proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain

Let $\alpha = a_1 + a_2 \sqrt{2}$ and $\beta = b_1 + b_2 \sqrt{2}$ be elements of $\mathbb{Z}[\sqrt{2}]$ with $\beta \neq 0$. We wish to show that there exist $\gamma$ and $\delta$ in $\mathbb{Z}[\sqrt{2}]$ such that $\alpha = \gamma\beta + \delta$ and $N(\delta) < N(\beta)$. To that end, note that in $\mathbb{Q}(\sqrt{2})$ we have $\frac{\alpha}{\beta} = c_1 + c_2 \sqrt{2}$, where $c_1 = \dfrac{a_1b_1 - 2a_2b_2}{b_1^2 - 2b_2^2}$ and $c_2 = \dfrac{a_2b_1 - a_1b_2}{b_1^2 - 2b_2^2}$.

Let $q_1$ be an integer closest to $c_1$ and $q_2$ an integer closest to $c_2$; then $|c_1 - q_1| \leq 1/2$ and $|c_2 - q_2| \leq 1/2$. Now let $\gamma = q_1 + q_2 \sqrt{2}$; certainly $\gamma \in \mathbb{Z}[\sqrt{2}]$. Next, let $\theta = (c_1 - q_1) + (c_2 - q_2) \sqrt{2}$. We have $\theta = \frac{\alpha}{\beta} - \gamma$, so that $\theta\beta = \alpha - \gamma\beta$.

Letting $\delta = \theta\beta$, we have $\alpha = \gamma\beta + \delta$. It remains to be shown that $N(\delta) < N(\beta)$. To that end, note that $$N(\theta) = |(c_1 - q_1)^2 - 2(c_2 - q_2)^2| \leq |(c_1 - q_1)^2| + |-2(c_2 - q_2)^2|$$ by the triangle inequality. Thus we have $$N(\theta) \leq (c_1 - q_1)^2 + 2(c_2 - q_2)^2 \leq (1/2)^2 + 2(1/2)^2 = 3/4.$$ In particular, $N(\delta) \leq \frac{3}{4}N(\beta)$ as desired.

$$\nu (a + b\sqrt{2} )\nu (c+d\sqrt{2} ) = |[(a + b\sqrt{2})(c + d\sqrt{2})][(a - b\sqrt{2})(c - d\sqrt{2})]|\\ =|(ac+2bd)+(ad+bc)\sqrt 2||ac+2bd-(ad+bc)\sqrt{2}|\\ =\nu(ac+2bd+(ad+bc)\sqrt{2})\\ =\nu((a+b\sqrt2)(c+d\sqrt2)) $$

Alternatively, using Jyrki's comment and noting that $\phi(a+b\sqrt 2)=a-b\sqrt 2$ is a ring homomorphism, we have $\nu(xy)=|xy\phi(xy)|=|x\phi(x)||y\phi(y)|=\nu(x)\nu(y)$.