Proof.
It is clear that (1) implies (2). Conversely, assume (2) and let $\mathcal{F}$ be a torsion abelian sheaf on $X_{\acute{e}tale}$. Let $Y' \to Y$ be a morphism of schemes and let $X' = Y' \times _ Y X$ with projections $g' : X' \to X$ and $f' : X' \to Y'$ as in diagram (59.91.5.1). We want to show the maps of sheaves
\[ g^{-1}R^ qf_*\mathcal{F} \longrightarrow R^ qf'_*(g')^{-1}\mathcal{F} \]
are isomorphisms for all $q \geq 0$.
For every $n \geq 1$, let $\mathcal{F}[n]$ be the subsheaf of sections of $\mathcal{F}$ annihilated by $n$. Then $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}[n]$. The functors $g^{-1}$ and $(g')^{-1}$ commute with arbitrary colimits (as left adjoints). Taking higher direct images along $f$ or $f'$ commutes with filtered colimits by Lemma 59.51.7. Hence we see that
\[ g^{-1}R^ qf_*\mathcal{F} = \mathop{\mathrm{colim}}\nolimits g^{-1}R^ qf_*\mathcal{F}[n] \quad \text{and}\quad R^ qf'_*(g')^{-1}\mathcal{F} = \mathop{\mathrm{colim}}\nolimits R^ qf'_*(g')^{-1}\mathcal{F}[n] \]
Thus it suffices to prove the result in case $\mathcal{F}$ is annihilated by a positive integer $n$.
If $n = \ell n'$ for some prime number $\ell $, then we obtain a short exact sequence
\[ 0 \to \mathcal{F}[\ell ] \to \mathcal{F} \to \mathcal{F}/\mathcal{F}[\ell ] \to 0 \]
Observe that $\mathcal{F}/\mathcal{F}[\ell ]$ is annihilated by $n'$. Moreover, if the result holds for both $\mathcal{F}[\ell ]$ and $\mathcal{F}/\mathcal{F}[\ell ]$, then the result holds by the long exact sequence of higher direct images (and the $5$ lemma). In this way we reduce to the case that $\mathcal{F}$ is annihilated by a prime number $\ell $.
Assume $\mathcal{F}$ is annihilated by a prime number $\ell $. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $ in $D(X_{\acute{e}tale}, \mathbf{Z}/\ell \mathbf{Z})$. Applying assumption (2) and Leray's acyclicity lemma (Derived Categories, Lemma 13.16.7) we see that
\[ f'_*(g')^{-1}\mathcal{I}^\bullet \]
computes $Rf'_*(g')^{-1}\mathcal{F}$. We conclude by applying Lemma 59.91.5.
$\square$
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