Hint $\ $ It is an immediate consequence of the following more general lemma, which is the basis of a general result on linear independence of square roots due to Besicovitch (see below).
Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $ all are not in $\rm\:K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\:K$
Proof $\ \ $ Let $\rm\ L = K(\sqrt{b}).\,$ $\rm\, [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K,\,$ so it is suffices to prove $\rm\: [L(\sqrt{a}):L] = 2.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b}).\, $ Then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K.\:$ But that is impossible since squaring yields $\ \rm\color{#c00}{(1)}:\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ contra hypotheses, as follows
$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $\,\color{#c00}{(1)}\,$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$
$\rm\qquad\qquad s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r \in K$
$\rm\qquad\qquad r = 0\ \ \Rightarrow\ \ \sqrt{a\:b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b},\: $ times $\rm\:\sqrt{b}\quad\,$ QED
Remark $\ $ Using the above as the inductive step one easily proves the following
Theorem $\ $ Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$ square roots $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts $\rm\ a,\:b,\:\ldots \in Q\:.\:$ If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n.\:$ Hence the $\rm\:2^n\:$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm\:L\:$ over $\rm\:Q.$