Find all intermediate fields of extension $mathbb{Q}(sqrt{2} , sqrt{3}) : mathbb{Q}$ without using Galois correspondence.

I'm in the middle of some notes which claim it should be possible to show that all the intermediate fields of the extension $\mathbb{Q}(\sqrt{2} , \sqrt{3}) : \mathbb{Q}$ are -

$\mathbb{Q}(\sqrt{2}, \sqrt{3}), \mathbb{Q}, \mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{3}), \mathbb{Q}(\sqrt{6})$.

But this is before covering the Galois correspondence so we can't just find all subgroups of $\operatorname {Gal}(\mathbb{Q}(\sqrt{2} , \sqrt{3}) : \mathbb{Q})$ and count the number of subgroups to tell us there can be no more intermediate fields.

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For example, one can show that the only intermediate fields of $\mathbb{Q}(\sqrt{2}) : \mathbb{Q}$ are $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}$ as follows:

By the Tower law, for an intermediate field $K$ where $\mathbb{Q} \subseteq K \subseteq \mathbb{Q}(\sqrt{2})$,

$|\mathbb{Q}(\sqrt{2}) : \mathbb{Q}| = |\mathbb{Q}(\sqrt{2}) : K| \cdot |K : \mathbb{Q}| = \deg(x^2 - 2) = 2$ as $x^2 - 2$ is irreducible over $\mathbb{Q}$.

Thus either $|\mathbb{Q}(\sqrt{2}) : K| = 1$, in which case $K=\mathbb{Q}(\sqrt{2})$

or

$|K : \mathbb{Q}| = 1$, in which case $K=\mathbb{Q}$

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Using a similar method, I can get as far as showing that, since $|\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}| = |\mathbb{Q}(\sqrt{2}, \sqrt{3}) : K | \cdot |K : \mathbb{Q}| = 4$ for an intermediate field $K$,

either $|\mathbb{Q}(\sqrt{2}, \sqrt{3}) : K | = 1$, in which case $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$

or $|K : \mathbb{Q}| = 1$, in which case $K = \mathbb{Q}$

or $|\mathbb{Q}(\sqrt{2}, \sqrt{3}) : K | = |K : \mathbb{Q}| = 2$

which is where I get stuck...

Any pointers appreciated!