[ This is question that appears as an excercise in Ian Stewart's book Galois Theory (4th edition). ]
I think the way to go about it is as follows, would appreciate someone's input on whether I got this right:
Write $L = \mathbb{Q}(\sqrt{5},\sqrt{7})$
We'll prove that for $L' = \mathbb{Q}(\sqrt{5} + \sqrt{7})$, it holds that both $L \subseteq L'$ as well as $L' \subseteq L$, hence $L = L'$.
The direction $L' \subseteq L$ holds trivially.
For the other direction we have the following (using basic closure properties of fields and the binomial formula for n=3):
$(\sqrt{5} + \sqrt{7})^3 = 5\sqrt{5} + 3\cdot5\sqrt{7} + 3\cdot7\sqrt{5} + 7\sqrt{7} = 26\sqrt{5} + 22\sqrt{7} \in L'$
hence
$(26\sqrt{5} + 22\sqrt{7}) - 22\cdot(\sqrt{5} + \sqrt{7}) \in L' \implies 4\sqrt{5} \in L' \implies \sqrt{5} \in L'$
Therefore, again by basic field properties, $(\sqrt{5} + \sqrt{7}) - \sqrt{5} = \sqrt{7} \in L'$
QED
Is this kosher? I'm still not very comfortable with field extensions, any input appreciated!