Proof. We have that \mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a,b \in \mathbb{Q} \}. So we need to show that all elements in \mathbb{Q}[\sqrt{2}] do have an inverse. Indeed, if we take arbitrary a,b \in \mathbb{Q} such that a + b \sqrt{2} \in \mathbb{Q}[\sqrt{2}], then
\begin{align*}
1 &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a + b \sqrt{2})(a - b \sqrt{2})} \\
&= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a^2 - ab\sqrt{2} + ab\sqrt{2} - 2b^2)} \\
&= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a^2 - 2b^2)}
\end{align*}