Theorem   Let $m,n \in S$ with $m \neq n$ Then $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ are not isomorphic.

Proof. Suppose that $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ are isomorphic. Let $\varphi \colon \mathbb{Q}(\sqrt{m}) \to \mathbb{Q}(\sqrt{n})$ be a field isomorphism. Recall that field homomorphisms fix prime subfields. Thus, for every $x \in \mathbb{Q}$ $\varphi(x)=x$

Let $a,b \in \mathbb{Q}$ with $\varphi(\sqrt{m})=a+b\sqrt{n}$ Since $\varphi(a)=a$ and $\varphi$ is injective, $b \neq 0$ Also, $m=\varphi(m)=\varphi((\sqrt{m})^2)=(\varphi(\sqrt{m}))^2=(a+b\sqrt{n})^2=a^2+2ab\sqrt{n}+b^2n$ If $a \neq 0$ then $\displaystyle \sqrt{n}=\frac{m-a^2-b^2n}{2ab} \in \mathbb{Q}$ a contradiction. Thus, $a=0$ Therefore, $m=b^2n$ Since $m$ is squarefree, $b^2=1$ Hence, $m=n$ a contradiction. It follows that $K$ and $L$ are not isomorphic.

Corollary   There are infinitely many distinct quadratic fields.

Proof. Note that there are infinitely many elements of $S$ Moreover, if $m$ and $n$ are distinct elements of $S$ then $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ are not isomorphic and thus cannot be equal.