Show that $\mathbb Q(\sqrt m,\sqrt n)=\mathbb Q(\sqrt {m}+\sqrt {n})$. Show that $\mathbb Q(\sqrt m,\sqrt n)=\mathbb Q(\sqrt {m}+\sqrt {n})$ 
My attempt: It is obvious that $\mathbb Q(\sqrt {m}+\sqrt {n}) \subset \mathbb Q(\sqrt m,\sqrt n) $ .
Is this proof is correct?

It is good, but you can shorten it to just deduce that
$$
2(m-n)\sqrt{n}\in\mathbb{Q}(\sqrt{m}+\sqrt{n})
$$
If $m=n$ the statement $\mathbb{Q}(\sqrt{m}+\sqrt{n})=\mathbb{Q}(\sqrt{m},\sqrt{n})$ is obvious, so we can assume $m\ne n$. Thus $\sqrt{n}\in\mathbb{Q}(\sqrt{m}+\sqrt{n})$ and so also
$$
\sqrt{m}=(\sqrt{m}+\sqrt{n})-\sqrt{n}\in\mathbb{Q}(\sqrt{m}+\sqrt{n})
$$
