We first use the induction principle to prove $i_X$ is fully faithful. More precisely, we will use Lemma \ref{lemma-induction-principle-enlarge}. Let $(U \subset W, V \to W)$ be an elementary distinguished square with $V$ affine and $U, W$ quasi-compact open in $X$. Assume that $i_U$ is fully faithful. We have to show that $i_W$ is fully faithful. We may replace $X$ by $W$, i.e., we may assume $W = X$ (we do this just to simplify the notation -- observe that the condition in the statement of the lemma is preserved under this operation). \medskip\noindent Suppose that $A, B$ are objects of $D(\QCoh(\mathcal{O}_X))$. We want to show that $$ \Hom_{D(\QCoh(\mathcal{O}_X))}(A, B) \longrightarrow \Hom_{D(\mathcal{O}_X)}(i_X(A), i_X(B)) $$ is bijective. Let $T = |X| \setminus |U|$. \medskip\noindent Assume first $i_X(B)$ is supported on $T$. In this case the map $$ i_X(B) \to Rj_{V, *}(i_X(B)|_V) = Rj_{V, *}(i_V(B|_V)) $$ is a quasi-isomorphism (Lemma \ref{lemma-pushforward-with-support-in-open}). By assumption we have an isomorphism $i_X(\Phi(B|_V)) \to Rj_{V, *}(i_V(B|_V))$ in $D(\mathcal{O}_X)$. Moreover, $\Phi$ and ${-}|_V$ are adjoint functors on the derived categories of quasi-coherent modules (by Derived Categories, Lemma \ref{derived-lemma-derived-adjoint-functors}). The adjunction map $B \to \Phi(B|_V)$ becomes an isomorphism after applying $i_X$, whence is an isomorphism in $D(\QCoh(\mathcal{O}_X))$. Hence \begin{align*} \Mor_{D(\QCoh(\mathcal{O}_X))}(A, B) & = \Mor_{D(\QCoh(\mathcal{O}_X))}(A, \Phi(B|_V)) \\ & = \Mor_{D(\QCoh(\mathcal{O}_V))}(A|_V, B|_V) \\ & = \Mor_{D(\mathcal{O}_V)}(i_V(A|_V), i_V(B|_V)) \\ & = \Mor_{D(\mathcal{O}_X)}(i_X(A), Rj_{V, *}(i_V(B|_V))) \\ & = \Mor_{D(\mathcal{O}_X)}(i_X(A), i_X(B)) \end{align*} as desired. Here we have used that $i_V$ is fully faithful (Lemma \ref{lemma-affine-coherator}). \medskip\noindent In general, choose any complex $\mathcal{B}^\bullet$ of quasi-coherent $\mathcal{O}_X$-modules representing $B$. Next, choose any quasi-isomorphism $s : \mathcal{B}^\bullet|_U \to \mathcal{C}^\bullet$ of complexes of quasi-coherent modules on $U$. As $j_U : U \to X$ is quasi-compact and quasi-separated the functor $j_{U, *}$ transforms quasi-coherent modules into quasi-coherent modules (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-pushforward}). Thus there is a canonical map $\mathcal{B}^\bullet \to j_{U, *}(\mathcal{B}^\bullet|_U) \to j_{U, *}\mathcal{C}^\bullet$ of complexes of quasi-coherent modules on $X$. Set $B'' = j_{U, *}\mathcal{C}^\bullet$ in $D(\QCoh(\mathcal{O}_X))$ and choose a distinguished triangle $$ B \to B'' \to B' \to B[1] $$ in $D(\QCoh(\mathcal{O}_X))$. Since the first arrow of the triangle restricts to an isomorphism over $U$ we see that $B'$ is supported on $T$. Hence in the diagram $$ \xymatrix{ \Hom_{D(\QCoh(\mathcal{O}_X))}(A, B'[-1]) \ar[r] \ar[d] & \Hom_{D(\mathcal{O}_X)}(i_X(A), i_X(B')[-1]) \ar[d] \\ \Hom_{D(\QCoh(\mathcal{O}_X))}(A, B) \ar[r] \ar[d] & \Hom_{D(\mathcal{O}_X)}(i_X(A), i_X(B)) \ar[d] \\ \Hom_{D(\QCoh(\mathcal{O}_X))}(A, B'') \ar[r] \ar[d] & \Hom_{D(\mathcal{O}_X)}(i_X(A), i_X(B'')) \ar[d] \\ \Hom_{D(\QCoh(\mathcal{O}_X))}(A, B') \ar[r] & \Hom_{D(\mathcal{O}_X)}(i_X(A), i_X(B')) } $$ we have exact columns and the top and bottom horizontal arrows are bijective. Finally, choose a complex $\mathcal{A}^\bullet$ of quasi-coherent modules representing $A$. \medskip\noindent Let $\alpha : i_X(A) \to i_X(B)$ be a morphism between in $D(\mathcal{O}_X)$. The restriction $\alpha|_U$ comes from a morphism in $D(\QCoh(\mathcal{O}_U))$ as $i_U$ is fully faithful. Hence there exists a choice of $s : \mathcal{B}^\bullet|_U \to \mathcal{C}^\bullet$ as above such that $\alpha|_U$ is represented by an actual map of complexes $\mathcal{A}^\bullet|_U \to \mathcal{C}^\bullet$. This corresponds to a map of complexes $\mathcal{A} \to j_{U, *}\mathcal{C}^\bullet$. In other words, the image of $\alpha$ in $\Hom_{D(\mathcal{O}_X)}(i_X(A), i_X(B''))$ comes from an element of $\Hom_{D(\QCoh(\mathcal{O}_X))}(A, B'')$. A diagram chase then shows that $\alpha$ comes from a morphism $A \to B$ in $D(\QCoh(\mathcal{O}_X))$. Finally, suppose that $a : A \to B$ is a morphism of $D(\QCoh(\mathcal{O}_X))$ which becomes zero in $D(\mathcal{O}_X)$. After choosing $\mathcal{B}^\bullet$ suitably, we may assume $a$ is represented by a morphism of complexes $a^\bullet : \mathcal{A}^\bullet \to \mathcal{B}^\bullet$. Since $i_U$ is fully faithul the restriction $a^\bullet|_U$ is zero in $D(\QCoh(\mathcal{O}_U))$. Thus we can choose $s$ such that $s \circ a^\bullet|_U : \mathcal{A}^\bullet|_U \to \mathcal{C}^\bullet$ is homotopic to zero. Applying the functor $j_{U, *}$ we conclude that $\mathcal{A}^\bullet \to j_{U, *}\mathcal{C}^\bullet$ is homotopic to zero. Thus $a$ maps to zero in $\Hom_{D(\QCoh(\mathcal{O}_X))}(A, B'')$. Thus we may assume that $a$ is the image of an element of $b \in \Hom_{D(\QCoh(\mathcal{O}_X))}(A, B'[-1])$. The image of $b$ in $\Hom_{D(\mathcal{O}_X)}(i_X(A), i_X(B')[-1])$ comes from a $\gamma \in \Hom_{D(\mathcal{O}_X)}(A, B''[-1])$ (as $a$ maps to zero in the group on the right). Since we've seen above the horizontal arrows are surjective, we see that $\gamma$ comes from a $c$ in $\Hom_{D(\QCoh(\mathcal{O}_X))}(A, B''[-1])$ which implies $a = 0$ as desired. \medskip\noindent At this point we know that $i_X$ is fully faithful for our original $X$. Since $RQ_X$ is its right adjoint, we see that $RQ_X \circ i_X = \text{id}$ (Categories, Lemma \ref{categories-lemma-adjoint-fully-faithful}). To finish the proof we show that for any $E$ in $D_\QCoh(\mathcal{O}_X)$ the map $i_X(RQ_X(E)) \to E$ is an isomorphism. Choose a distinguished triangle $$ i_X(RQ_X(E)) \to E \to E' \to i_X(RQ_X(E))[1] $$ in $D_\QCoh(\mathcal{O}_X)$. A formal argument using the above shows that $i_X(RQ_X(E')) = 0$. Thus it suffices to prove that for $E \in D_\QCoh(\mathcal{O}_X)$ the condition $i_X(RQ_X(E)) = 0$ implies that $E = 0$. Consider an \'etale morphism $j : V \to X$ with $V$ affine. By Lemmas \ref{lemma-affine-coherator} and \ref{lemma-flat-pushforward-coherator} and our assumption we have $$ Rj_*(E|_V) = Rj_*(i_V(RQ_V(E|_V))) = i_X(\Phi(RQ_V(E|_V))) = i_X(RQ_X(Rj_*(E|_V))) $$ Choose a distinguished triangle $$ E \to Rj_*(E|_V) \to E' \to E[1] $$ Apply $RQ_X$ to get a distinguished triangle $$ 0 \to RQ_X(Rj_*(E|_V)) \to RQ_X(E') \to 0[1] $$ in other words the map in the middle is an isomorphism. Combined with the string of equalities above we find that our first distinguished triangle becomes a distinguished triangle $$ E \to i_X(RQ_X(E')) \to E' \to E[1] $$ where the middle morphism is the adjunction map. However, the composition $E \to E'$ is zero, hence $E \to i_X(RQ_X(E'))$ is zero by adjunction! Since this morphism is isomorphic to the morphism $E \to Rj_*(E|_V)$ adjoint to $\text{id} : E|_V \to E|_V$ we conclude that $E|_V$ is zero. Since this holds for all affine $V$ \'etale over $X$ we conclude $E$ is zero as desired.