**Proof.**
By Properties of Spaces, Lemma 66.22.5 every algebraic space $Y$ étale over $X$ has dimension $\leq d$. If $Y$ is quasi-separated, the dimension of $Y$ is equal to the Krull dimension of $|Y|$ by Decent Spaces, Lemma 68.12.5. Also, if $Y$ is a scheme, then étale cohomology of $\mathcal{F}$ over $Y$, resp. étale cohomology of $\mathcal{F}$ with support in a closed subscheme, agrees with usual cohomology of $\mathcal{F}$, resp. usual cohomology with support in the closed subscheme. See Descent, Proposition 35.9.3 and Étale Cohomology, Lemma 59.79.5. We will use these facts without further mention.

By Decent Spaces, Lemma 68.8.6 there exist an integer $n$ and open subspaces

\[ \emptyset = U_{n + 1} \subset U_ n \subset U_{n - 1} \subset \ldots \subset U_1 = X \]

with the following property: setting $T_ p = U_ p \setminus U_{p + 1}$ (with reduced induced subspace structure) there exists a quasi-compact separated scheme $V_ p$ and a surjective étale morphism $f_ p : V_ p \to U_ p$ such that $f_ p^{-1}(T_ p) \to T_ p$ is an isomorphism.

As $U_ n = V_ n$ is a scheme, our initial remarks imply the cohomology of $\mathcal{F}$ over $U_ n$ vanishes in degrees $> d$ by Cohomology, Proposition 20.22.4. Suppose we have shown, by induction, that $H^ q(U_{p + 1}, \mathcal{F}|_{U_{p + 1}}) = 0$ for $q > d$. It suffices to show $H_{T_ p}^ q(U_ p, \mathcal{F})$ for $q > d$ is zero in order to conclude the vanishing of cohomology of $\mathcal{F}$ over $U_ p$ in degrees $> d$. However, we have

\[ H^ q_{T_ p}(U_ p, \mathcal{F}) = H^ q_{f_ p^{-1}(T_ p)}(V_ p, \mathcal{F}) \]

by Lemma 69.9.3 and as $V_ p$ is a scheme we obtain the desired vanishing from Cohomology, Proposition 20.22.4. In this way we conclude that (1) is true.

To prove (2) let $U \subset X$ be a quasi-compact open subspace. Consider the open subspace $U' = U \cup U_ n$. Let $Z = U' \setminus U$. Then $g : U_ n \to U'$ is an étale morphism such that $g^{-1}(Z) \to Z$ is an isomorphism. Hence by Lemma 69.9.3 we have $H^ q_ Z(U', \mathcal{F}) = H^ q_ Z(U_ n, \mathcal{F})$ which vanishes in degree $> d$ because $U_ n$ is a scheme and we can apply Cohomology, Proposition 20.22.4. We conclude that $H^ d(U', \mathcal{F}) \to H^ d(U, \mathcal{F})$ is surjective. Assume, by induction, that we have reduced our problem to the case where $U$ contains $U_{p + 1}$. Then we set $U' = U \cup U_ p$, set $Z = U' \setminus U$, and we argue using the morphism $f_ p : V_ p \to U'$ which is étale and has the property that $f_ p^{-1}(Z) \to Z$ is an isomorphism. In other words, we again see that

\[ H^ q_ Z(U', \mathcal{F}) = H^ q_{f_ p^{-1}(Z)}(V_ p, \mathcal{F}) \]

and we again see this vanishes in degrees $> d$. We conclude that $H^ d(U', \mathcal{F}) \to H^ d(U, \mathcal{F})$ is surjective. Eventually we reach the stage where $U_1 = X \subset U$ which finishes the proof.

A formal argument shows that (2) implies (3).
$\square$

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