By Lemma \ref{lemma-pseudo-coherent} we see that $E$ is an object of $D_\QCoh(\mathcal{O}_U)$. By Lemma \ref{lemma-lift-quasi-coherent} we may assume $E = E'|U$ for some object $E'$ of $D_\QCoh(\mathcal{O}_X)$. Write $X = \Spec(A)$. By Lemma \ref{lemma-affine-compare-bounded} we can find a complex $M^\bullet$ of $A$-modules whose associated complex of $\mathcal{O}_X$-modules is a representative of $E'$. \medskip\noindent Choose $f_1, \ldots, f_r \in A$ such that $U = D(f_1) \cup \ldots \cup D(f_r)$. By Lemma \ref{lemma-pseudo-coherent-affine} the complexes $M^\bullet_{f_j}$ are pseudo-coherent complexes of $A_{f_j}$-modules. Let $n$ be an integer. Assume we have a map of complexes $\alpha : F^\bullet \to M^\bullet$ where $F^\bullet$ is bounded above, $F^i = 0$ for $i < n$, each $F^i$ is a finite free $R$-module, such that $$ H^i(\alpha_{f_j}) : H^i(F^\bullet_{f_j}) \to H^i(M^\bullet_{f_j}) $$ is an isomorphism for $i > n$ and surjective for $i = n$. Picture $$ \xymatrix{ & F^n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ M^{n-1} \ar[r] & M^n \ar[r] & M^{n + 1} \ar[r] & \ldots } $$ Since each $M^\bullet_{f_j}$ has vanishing cohomology in large degrees we can find such a map for $n \gg 0$. By induction on $n$ we are going to extend this to a map of complexes $F^\bullet \to M^\bullet$ such that $H^i(\alpha_{f_j})$ is an isomorphism for all $i$. The lemma will follow by taking $\widetilde{F^\bullet}$. \medskip\noindent The induction step will be to extend the diagram above by adding $F^{n - 1}$. Let $C^\bullet$ be the cone on $\alpha$ (Derived Categories, Definition \ref{derived-definition-cone}). The long exact sequence of cohomology shows that $H^i(C^\bullet_{f_j}) = 0$ for $i \geq n$. By More on Algebra, Lemma \ref{more-algebra-lemma-cone-pseudo-coherent} we see that $C^\bullet_{f_j}$ is $(n - 1)$-pseudo-coherent. By More on Algebra, Lemma \ref{more-algebra-lemma-finite-cohomology} we see that $H^{-1}(C^\bullet_{f_j})$ is a finite $A_{f_j}$-module. Choose a finite free $A$-module $F^{n - 1}$ and an $A$-module $\beta : F^{n - 1} \to C^{-1}$ such that the composition $F^{n - 1} \to C^{n - 1} \to C^n$ is zero and such that $F^{n - 1}_{f_j}$ surjects onto $H^{n - 1}(C^\bullet_{f_j})$. (Some details omitted; hint: clear denominators.) Since $C^{n - 1} = M^{n - 1} \oplus F^n$ we can write $\beta = (\alpha^{n - 1}, -d^{n - 1})$. The vanishing of the composition $F^{n - 1} \to C^{n - 1} \to C^n$ implies these maps fit into a morphism of complexes $$ \xymatrix{ & F^{n - 1} \ar[d]^{\alpha^{n - 1}} \ar[r]_{d^{n - 1}} & F^n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & M^{n - 1} \ar[r] & M^n \ar[r] & M^{n + 1} \ar[r] & \ldots } $$ Moreover, these maps define a morphism of distinguished triangles $$ \xymatrix{ (F^n \to \ldots) \ar[r] \ar[d] & (F^{n-1} \to \ldots) \ar[r] \ar[d] & F^{n-1} \ar[r] \ar[d]_\beta & (F^n \to \ldots)[1] \ar[d] \\ (F^n \to \ldots) \ar[r] & M^\bullet \ar[r] & C^\bullet \ar[r] & (F^n \to \ldots)[1] } $$ Hence our choice of $\beta$ implies that the map of complexes $(F^{-1} \to \ldots) \to M^\bullet$ induces an isomorphism on cohomology localized at $f_j$ in degrees $\geq n$ and a surjection in degree $-1$. This finishes the proof of the lemma.