How to prove that $H^q(X_{et},F) \cong \prod_{i=1}^n H^q(X_{i_{et}}, F|_{X_i})$ for $X = X_1 \sqcup ... \sqcup X_n$

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Let $X = X_1 \sqcup ... \sqcup X_n$ be a disjoint union of schemes, and let $F \in \text{Ab}(X_{et})$ be an abelian sheaf on an etale site of $X$. I need to show that $H^q(X_{et},F) \cong \prod_{i=1}^n H^q(X_{i_{et}}, F|_{X_i}) $.

I am completely stuck on this one. Any help would be very appreciated. Thanks.

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There are plenty of way to do this.

The quickest is to use the Cech to derived spectral sequence. It reads $$ E_2^{pq}=\check{H}^p(\mathcal{U},\underline{H}^q(F))\Rightarrow H^{p+q}(X,F) $$ If $\mathcal{U}=\{X_i\}$ is a covering of $X$ by disjoint open subset, then $\check{H}^p(\mathcal{U},\cdot)=0$ for $p>0$ since there is no intersections. Thus the spectral sequence collapse and $$H^q(X,F)=\check{H}^0(\mathcal{U},\underline{H}^q(F))=\prod \underline{H}^q(F)(X_i)=\prod H^q(X_i,F_{|X_i})$$

You can also directly show if $j_i:X_i\rightarrow X$ denotes the inclusion, then the functors $j_i^*,{j_i}_*$ are adjoint in both direction, that is $j_i^*$ is a left (as usual) but also a right adjoint of ${j_i}_*$. In particular, they are both exact, preserving injective (and projective), so you can quickly see that the derived functor of $\Gamma(X,F)=\prod\Gamma(X_i,j_i^*F)$ is $\prod H^p(X_i,j_i^*F)$.

In fact, you even have an equivalence of categories $Sh(X)\simeq \prod Sh(X_i)$.