We prove this result by induction on $d$. \medskip\noindent If $d = 0$, then $X$ is a profinite space, see Topology, Lemma \ref{topology-lemma-characterize-profinite-spectral}. Thus (1) holds by Lemma \ref{lemma-vanishing-for-profinite}. If $U \subset X$ is quasi-compact open, then $U$ is also closed as a quasi-compact subset of a Hausdorff space. Hence $X = U \amalg (X \setminus U)$ as a topological space and we see that (2) holds. Given $Z$ as in (3) we consider the long exact sequence $$ H^{q - 1}(X, \mathcal{F}) \to H^{q - 1}(X \setminus Z, \mathcal{F}) \to H^q_Z(X, \mathcal{F}) \to H^q(X, \mathcal{F}) $$ Since $X$ and $U = X \setminus Z$ are profinite (namely $U$ is quasi-compact because $Z$ is constructible) and since we have (2) and (1) we obtain the desired vanishing of the cohomology groups with support in $Z$. \medskip\noindent Induction step. Assume $d \geq 1$ and assume the proposition is valid for all spectral spaces of dimension $< d$. We first prove part (2) for $X$. Let $U$ be a quasi-compact open. Let $\xi \in H^d(U, \mathcal{F})$. Set $Z = X \setminus U$. Let $W \subset X$ be the set of points specializing to $Z$. By Lemma \ref{lemma-cohomology-of-neighbourhoods-of-closed} we have $$ H^d(W \setminus Z, \mathcal{F}|_{W \setminus Z}) = \colim_{Z \subset V} H^d(V \setminus Z, \mathcal{F}) $$ where the colimit is over the quasi-compact open neighbourhoods $V$ of $Z$ in $X$. By Topology, Lemma \ref{topology-lemma-make-spectral-space} we see that $W \setminus Z$ is a spectral space. Since every point of $W$ specializes to a point of $Z$, we see that $W \setminus Z$ is a spectral space of Krull dimension $< d$. By induction hypothesis we see that the image of $\xi$ in $H^d(W \setminus Z, \mathcal{F}|_{W \setminus Z})$ is zero. By the displayed formula, there exists a $Z \subset V \subset X$ quasi-compact open such that $\xi|_{V \setminus Z} = 0$. Since $V \setminus Z = V \cap U$ we conclude by the Mayer-Vietoris (Lemma \ref{lemma-mayer-vietoris}) for the covering $X = U \cup V$ that there exists a $\tilde \xi \in H^d(X, \mathcal{F})$ which restricts to $\xi$ on $U$ and to zero on $V$. In other words, part (2) is true. \medskip\noindent Proof of part (1) assuming (2). Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$. Set $$ \mathcal{G} = \Im(\mathcal{I}^{d - 1} \to \mathcal{I}^d) = \Ker(\mathcal{I}^d \to \mathcal{I}^{d + 1}) $$ For $U \subset X$ quasi-compact open we have a map of exact sequences as follows $$ \xymatrix{ \mathcal{I}^{d - 1}(X) \ar[r] \ar[d] & \mathcal{G}(X) \ar[r] \ar[d] & H^d(X, \mathcal{F}) \ar[d] \ar[r] & 0 \\ \mathcal{I}^{d - 1}(U) \ar[r] & \mathcal{G}(U) \ar[r] & H^d(U, \mathcal{F}) \ar[r] & 0 } $$ The sheaf $\mathcal{I}^{d - 1}$ is flasque by Lemma \ref{lemma-injective-flasque} and the fact that $d \geq 1$. By part (2) we see that the right vertical arrow is surjective. We conclude by a diagram chase that the map $\mathcal{G}(X) \to \mathcal{G}(U)$ is surjective. By Lemma \ref{lemma-vanishing-ravi} we conclude that $\check{H}^q(\mathcal{U}, \mathcal{G}) = 0$ for $q > 0$ and any finite covering $\mathcal{U} : U = U_1 \cup \ldots \cup U_n$ of a quasi-compact open by quasi-compact opens. Applying Lemma \ref{lemma-cech-vanish-basis} we find that $H^q(U, \mathcal{G}) = 0$ for all $q > 0$ and all quasi-compact opens $U$ of $X$. By Leray's acyclicity lemma (Derived Categories, Lemma \ref{derived-lemma-leray-acyclicity}) we conclude that $$ H^q(X, \mathcal{F}) = H^q\left( \Gamma(X, \mathcal{I}^0) \to \ldots \to \Gamma(X, \mathcal{I}^{d - 1}) \to \Gamma(X, \mathcal{G}) \right) $$ In particular the cohomology group vanishes if $q > d$. \medskip\noindent Proof of (3).  Given $Z$ as in (3) we consider the long exact sequence $$ H^{q - 1}(X, \mathcal{F}) \to H^{q - 1}(X \setminus Z, \mathcal{F}) \to H^q_Z(X, \mathcal{F}) \to H^q(X, \mathcal{F}) $$ Since $X$ and $U = X \setminus Z$ are spectral spaces (Topology, Lemma \ref{topology-lemma-spectral-sub}) of dimension $\leq d$ and since we have (2) and (1) we obtain the desired vanishing.