Before you answer, yes i am aware of the outcome of the solution by a "brute force" method of finding the basis for $\mathbb{Q}(\sqrt{2},\sqrt{3})$. I just want to know what the authors (Dummit/Foote pg 526) are talking about in their reasoning.
In the first sentence they claimed, the degree extension is $\leq 2$ because $[\mathbb{Q}(\sqrt{3}),\mathbb{Q}] = 2$.
We have $[\Bbb Q(\sqrt3):\Bbb Q]=2$. This means that there is an irreducible polynomial of degree $2$ with rational coefficients where $\sqrt3$ is a root. (One such polynomial is $x^2-3$, although the exact polynomial expression is of little importance..)
Now for $[\Bbb Q(\sqrt3,\sqrt2):\Bbb Q(\sqrt2)]$. It corresponds to the degree of an irreducible polynomial with coefficients in $\Bbb Q(\sqrt2)$ where $\sqrt3$ is a root. Clearly the polynomial from before is a polynomial with valid coefficients and the right root. Is it irreducible? Don't know, don't care. Regardless, there is an irreducible. polynomial of either first or second degree with $\sqrt3$ as a root, meaning the extension has either degree $1$ or degree $2$.