We want to show that $\mathbb{Q}[\sqrt{2}] \subsetneqq \mathbb{Q}[\sqrt{2}+\sqrt{3}]$.
Answer: We know that $\mathbb{Q}[\sqrt{2}]$ is the smallest integral domain which containts $\mathbb{Q}\cup\{\sqrt{2}\}$ & $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ is the smallest integral domain which containts $\mathbb{Q}\cup\{\sqrt{2}+\sqrt{3}\}$. So, for $``\subseteq"$, we have to proove that $\sqrt{2}\in \mathbb{Q}[\sqrt{2}+\sqrt{3}]$.
It's clear that $\sqrt{2}+\sqrt{3} \in \mathbb{Q}[\sqrt{2}+\sqrt{3}]$.
I have seen in a book that $$-\sqrt{2}+\sqrt{3}=\frac{1}{\sqrt{2}+\sqrt{3}} \in \mathbb{Q}[\sqrt{2}+\sqrt{3}]\ \ (*)$$ and I don't understand why this happens. Could you explain this please?
If this is valid, then we can see that $\sqrt{2}=\frac{1}{2}(\sqrt{2}+\sqrt{3})-\frac{1}{2}(-\sqrt{2}+\sqrt{3})\in \mathbb{Q}[\sqrt{2}+\sqrt{3}]$ and we are done.
Now for $``\neq".$ We suppose that $\sqrt{2}+\sqrt{3}\in \mathbb{Q}[\sqrt{2}]$. We know that $\mathbb{Q}[\sqrt{2}]$ is a $\mathbb{Q}$-vector space and $\{1,\sqrt{2}\}$ is one base. So, $$\sqrt{2}+\sqrt{3}\in \mathbb{Q}[\sqrt{2}]=\langle 1,\sqrt{2} \rangle \iff \exists \ a,b \in \mathbb{Q}:\sqrt{2}+\sqrt{3}=a+b\sqrt2.$$ We have the cases: If $a=b=0$, then $\sqrt{2}=-\sqrt{3}$, contradiction. If $a=0,b\neq0$, then $b=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}} \notin \mathbb{Q}$, contradiction. If $b=0,a\neq0$, then $a=\sqrt{2}+\sqrt{3}\notin \mathbb{Q}$, contradiction. If $ab\neq0$, then $\sqrt{3}=a+(b-1)\sqrt{2}\implies3=a^2+2a(b-1)\sqrt{2}+2(b-1)^2$. If $b=1$ then $a=\pm \sqrt{3}\notin \mathbb{Q}$, contradiction. If $b\neq1$, then $\frac{3-a^2-2(b-1)^2}{2a(b-1)}=\sqrt{2}$, contradiction. So, $\sqrt{2}+\sqrt{3}\notin \mathbb{Q}[\sqrt{2}]$, and finally $\mathbb{Q}[\sqrt{2}] \subsetneqq \mathbb{Q}[\sqrt{2}+\sqrt{3}]$.
I would like to explain me the $(*)$ relation and check the other part. Any other way is welcome. Thank you in advance.
An idea using your own work: you've already shown that $\;-\sqrt2+\sqrt3\in\Bbb Q(\sqrt2+\sqrt3)\;$ , so we also get that
$$2\sqrt3=(-\sqrt2+\sqrt3)+(\sqrt2+\sqrt3)\in\Bbb Q(\sqrt2+\sqrt3)\implies \sqrt3\in\Bbb Q(\sqrt2+\sqrt3)$$
Thus, if $\;\Bbb Q(\sqrt2)=\Bbb Q(\sqrt2+\sqrt3)\;$ , then $\;\sqrt3\in\Bbb Q(\sqrt2)\;$ , but then
$$\sqrt3=a+b\sqrt2\;,\;\;a,b\in\Bbb Q\implies 3=a^2+2b^2+2ab\sqrt2\implies$$
$$\sqrt2=\frac{3-a^2-2b^2}{2ab}\in\Bbb Q\;,\;\;\text{ contradiction}$$
I leave to you to check in the above the easy cases when $\;a=0\;$ or $\;b=0\;$ .