By Theorem \ref{theorem-K-injective-embedding-grothendieck} we obtain quasi-isomorphisms $i : K^\bullet \to I^\bullet$ and $i^p : F^pK^\bullet \to I^{p, \bullet}$ as well as commutative diagrams $$ \vcenter{ \xymatrix{ K^\bullet \ar[d]_i & F^pK^\bullet \ar[l] \ar[d]_{i^p} \\ I^\bullet & I^{p, \bullet} \ar[l]_{\alpha^p} } } \quad\text{and}\quad \vcenter{ \xymatrix{ F^{p'}K^\bullet \ar[d]_{i^{p'}} & F^pK^\bullet \ar[l] \ar[d]_{i^p} \\ I^{p', \bullet} & I^{p, \bullet} \ar[l]_{\alpha^{p p'}} } } \quad\text{for }p' \leq p $$ such that $\alpha^p \circ \alpha^{p' p} = \alpha^{p'}$ and $\alpha^{p'p''} \circ \alpha^{pp'} = \alpha^{pp''}$. The problem is that the maps $\alpha^p : I^{p, \bullet} \to I^\bullet$ need not be injective. For each $p$ we choose an injection $t^p : I^{p, \bullet} \to J^{p, \bullet}$ into an acyclic K-injective complex $J^{p, \bullet}$ whose terms are injective objects of $\mathcal{A}$ (first map to the cone on the identity and then use the theorem). Choose a map of complexes $s^p : I^\bullet \to J^{p, \bullet}$ such that the following diagram commutes $$ \xymatrix{ K^\bullet \ar[d]_i & F^pK^\bullet \ar[l] \ar[d]_{i^p} \\ I^\bullet \ar[rd]_{s^p} & I^{p, \bullet} \ar[d]^{t^p} \\ & J^{p, \bullet} } $$ This is possible: the composition $F^pK^\bullet \to J^{p, \bullet}$ is homotopic to zero because $J^{p, \bullet}$ is acyclic and K-injective (Derived Categories, Lemma \ref{derived-lemma-K-injective}). Since the objects $J^{p, n - 1}$ are injective and since $F^pK^n \to K^n \to I^n$ are injective morphisms, we can lift the maps $F^pK^n \to J^{p, n - 1}$ giving the homotopy to a map $h^n : I^n \to J^{p, n - 1}$. Then we set $s^p$ equal to $h \circ \text{d} + \text{d} \circ h$. (Warning: It will not be the case that $t^p = s^p \circ \alpha^p$, so we have to be careful not to use this below.) \medskip\noindent Consider $$ J^\bullet = I^\bullet \times \prod\nolimits_p J^{p, \bullet} $$ Because products in $D(\mathcal{A})$ are given by taking products of K-injective complexes (Lemma \ref{lemma-derived-products}) and since $J^{p, \bullet}$ is isomorphic to $0$ in $D(\mathcal{A})$ we see that $J^\bullet \to I^\bullet$ is an isomorphism in $D(\mathcal{A})$. Consider the map $$ j = i \times (s^p \circ i)_{p \in \mathbf{Z}} : K^\bullet \longrightarrow I^\bullet \times \prod\nolimits_p J^{p, \bullet} = J^\bullet $$ By our remarks above this is a quasi-isomorphism. It is also injective. For $p \in \mathbf{Z}$ we let $F^pJ^\bullet \subset J^\bullet$ be $$ \Im\left( \alpha^p \times (t^{p'} \circ \alpha^{pp'})_{p' \leq p} : I^{p, \bullet} \to I^\bullet \times \prod\nolimits_{p' \leq p} J^{p', \bullet} \right) \times \prod\nolimits_{p' > p} J^{p', \bullet} $$ This complex is isomorphic to the complex $I^{p, \bullet} \times \prod_{p' > p} J^{p, \bullet}$ as $\alpha^{pp} = \text{id}$ and $t^p$ is injective. Hence $F^pJ^\bullet$ is quasi-isomorphic to $I^{p, \bullet}$ (argue as above). We have $j(F^pK^\bullet) \subset F^pJ^\bullet$ because of the commutativity of the diagram above. The corresponding map of complexes $F^pK^\bullet \to F^pJ^\bullet$ is a quasi-isomorphism by what we just said. Finally, to see that $F^{p + 1}J^\bullet \subset F^pJ^\bullet$ use that $\alpha^{p + 1p} \circ \alpha^{pp'} = \alpha^{p + 1p'}$ and the commutativity of the first displayed diagram in the first paragraph of the proof. \medskip\noindent We claim that $j : K^\bullet \to J^\bullet$ is a solution to the problem posed by the lemma. Namely, $F^pJ^n$ is an injective object of $\mathcal{A}$ because it is isomorphic to $I^{p, n} \times \prod_{p' > p} J^{p', n}$ and products of injectives are injective. Then the injective map $F^pJ^n \to J^n$ splits and hence the quotient $J^n/F^pJ^n$ is injective as well as a direct summand of the injective object $J^n$. Similarly for $F^pJ^n/F^{p'}J^n$. This in particular means that $0 \to F^pJ^\bullet \to J^\bullet \to J^\bullet/F^pJ^\bullet \to 0$ is a termwise split short exact sequence of complexes, hence defines a distinguished triangle in $K(\mathcal{A})$ by fiat. Since $J^\bullet$ and $F^pJ^\bullet$ are K-injective complexes we see that the same is true for $J^\bullet/F^pJ^\bullet$ by Derived Categories, Lemma \ref{derived-lemma-triangle-K-injective}. A similar argument shows that $F^pJ^\bullet/F^{p'}J^\bullet$ is K-injective. By construction $j : K^\bullet \to J^\bullet$ and the induced maps $F^pK^\bullet \to F^pJ^\bullet$ are quasi-isomorphisms. Using the long exact cohomology sequences of the complexes in play we find that the same holds for $K^\bullet/F^pK^\bullet \to J^\bullet/F^pJ^\bullet$ and $F^pK^\bullet/F^{p'}K^\bullet \to F^pJ^\bullet/F^{p'}J^\bullet$.