In chapter 13 of Dummit and Foote, there is the following example:
Consider the field $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ generated over $\mathbb{Q}$ by $\sqrt{2}$ and $\sqrt{3}$. Since $\sqrt{3}$ is of degree 2 over $\mathbb{Q}$, the degree of the extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is at most 2...
I understand that $\sqrt{3}$ is of degree 2 over $\mathbb{Q}$ (it has minimal polynomial $x^2 - 3$), but I'm lost as to how this implies $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}(\sqrt{2})] \leq 2$.
I also know that $x^2 - 3 \in \mathbb{Q}(\sqrt{2})[x]$, so that the degree of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$ is at most 2. This would give $[\mathbb{Q}(\sqrt{3}): \mathbb{Q}(\sqrt{2})] \leq 2$. But I can't really see how this would imply $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}(\sqrt{2})] \leq 2$?
Your observation that $x^2-3\in\Bbb Q(\sqrt 2)][x]$ is the correct one. The reason it gives what you want is because either $x^2-3$ is irreducible, or it isn't. If it is, then $\Bbb Q(\sqrt 2)[x]/(x^2-3)$ is a field of degree $2$ over the base field because the ideal is maximal, giving degree $2$. If not, then it factors, and the only way is that it splits into a product of linear factors. But then this means there is already a root in $\Bbb Q(\sqrt 2)$ so $\Bbb Q(\sqrt 2, \sqrt 3)=\Bbb Q(\sqrt 2)$ and the degree is $1$. Either way it is at most $2$.