ok, there's an easier way. let $\displaystyle F$ be a subfield of $\displaystyle \mathbb{Q}(\sqrt{2})$. i showed in my previous post that $\displaystyle \mathbb{Q} \subseteq F.$ now

$\displaystyle 2=[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}):F][F:\mathbb{Q}] \ \ \ \ \ \ \ (*).$

so if $\displaystyle F \neq \mathbb{Q},$ then $\displaystyle [F:\mathbb{Q}] > 1$ and hence, by $\displaystyle (*)$, $\displaystyle [F:\mathbb{Q}]=2$ and $\displaystyle [\mathbb{Q}(\sqrt{2}):F]=1.$ thus $\displaystyle F=\mathbb{Q}(\sqrt{2}).$