As I said $\mathbb Q(\sqrt2)$ is the smallest field contianing $\mathbb Q$ and $\sqrt2$. This by definition! You have that $\mathbb Q(\sqrt2):=Frac(\mathbb Q[\sqrt2])$... but this is valid for every domain $R$; you can always consider $Frac(R)$ and by construction you can see there is a natural immersion $R\hookrightarrow Frac(R)$ given by $a\longmapsto\frac{a}{1_{R}}$ (keep in mind the construction you see in gettin' rational numbers from integers: you have an immersion $\mathbb Z\hookrightarrow\mathbb Q$ given by $h\longmapsto\frac{h}{1}$).
If you want you can think at $\mathbb Q(\sqrt2)$ as the intersection of all subfields of $\mathbb R$ containing $\mathbb Q$ and $\sqrt2$, i.e. $\mathbb Q(\sqrt2):=\bigcap_{F\leq\mathbb R \;s.t.\; \mathbb Q \subseteq F \;and\;\sqrt2\in F}F$, but this isn't useful for our purpose.
But if you want to see explicitly that $\mathbb Q[\sqrt2]\subseteq\mathbb Q(\sqrt2)$, having in mind that $\mathbb Q(\sqrt2)$ is by def. a field (the smallest, but now it's not relevant) containing both $\mathbb Q$ and $\sqrt2$, it should be clear that $q\in\mathbb Q(\sqrt2)\;\;\forall q\in\mathbb Q$, $\sqrt2\in\mathbb Q(\sqrt2)\Longrightarrow q\sqrt2\in\mathbb Q(\sqrt2)$ hence $q+p\sqrt2\in\mathbb Q(\sqrt2)$ for every $p,q\in\mathbb Q$. But these ones are all and only the elements of $\mathbb Q[\sqrt2]$, so we're saying exactly that
$\mathbb Q[\sqrt2]=\{q+p\sqrt2\;\ :\;q,p\in\mathbb Q\}\subseteq \mathbb Q(\sqrt2)$.
Pay attention! Before you prove that $\mathbb Q[\sqrt2]=\mathbb Q(\sqrt2)$ you can't know $\mathbb Q[\sqrt2]$ is a field... but you know this is a ring. Moreover $\mathbb Q(\sqrt2)$ is a field by definition. But every field is a ring, so what you have is an inclusion of rings!