Proof.
In case (1) the filtration on $T^\bullet $ is finite and the lemma follows immediately from Homology, Lemma 12.24.11. In case (2) choose $a < b$ such that $F^ iK^\bullet $ and $F^ jL^\bullet $ are acyclic for $i, j > b$ and $F^ iK^\bullet \to K^\bullet $ and $F^ jL^\bullet \to L^\bullet $ are quasi-isomorphisms for $i, j < a$. We claim that in this case the complex $F^ nT^\bullet $ is acyclic for $n > 2b$ and that $F^ nT^\bullet \to T^\bullet $ is a quasi-isomorphism for $n < 2a - 1$. Since the claim shows that Homology, Lemma 12.24.13 applies we see that our lemma is true.
Proof of the claim. Given $n$ and integers $i_1 \leq i_2$ consider
\[ S^{n, i_1, i_2} = \mathop{\mathrm{Im}}\left( \bigoplus \nolimits _{i_1 \leq i \leq i_2} \text{Tot}(F^ iK^\bullet \otimes _ R F^{n - i}L^\bullet ) \to \text{Tot}(K^\bullet \otimes _ R L^\bullet ) \right) \]
viewed as a subcomplex of $\text{Tot}(K^\bullet \otimes _ R L^\bullet )$. Observe that $F^ nT^\bullet = \mathop{\mathrm{colim}}\nolimits S^{n, i_1, i_2}$ is a filtered colimit. Thus to show the claim it suffices to show that $S^{n, i_1, i_2}$ is acyclic for $n > 2b$ and that $S^{n, i_1, i_2} \to \text{Tot}(K^\bullet \otimes L^\bullet )$ is a quasi-isomorphism for $n < 2a - 1$ and a cofinal set of choices of pairs $i_1, i_2$.
If $i_1 = i_2 = i$ and $n > 2b$, then
\[ S^{n, i, i} = \text{Tot}(F^ iK^\bullet \otimes _ R F^{n - i}L^\bullet ) \]
is acyclic: either $i > b$ and this complex is acyclic by our assumption that $F^{n - i}L^\bullet $ is K-flat, or $i \leq b$ and then $n - i \geq n - b > b$ and the same holds. Using our flatness assumptions the reader shows that there is a short exact sequence
\[ 0 \to S^{n + 1, i_2 + 1, i_2 + 1} \to S^{n, i_1, i_2} \oplus S^{n, i_2 + 1, i_2 + 1} \to S^{n, i_1, i_2 + 1} \to 0 \]
of complexes. Thus we see by induction on $i_2 - i_1$ that all of the complexes $S^{n, i_1, i_2}$ for $n > 2b$ are acyclic.
Assume $n < 2a - 1$. Observe that
\[ S^{n, a - 1, a - 1} = \text{Tot}(F^{a - 1}K^\bullet \otimes _ R F^{n - a + 1}L^\bullet ) \to \text{Tot}(K^\bullet \otimes _ R L^\bullet ) \]
is a quasi-isomorphism because both $a - 1 < a$ and $n - a + 1 < a$ and all our complexes are K-flat hence both sides compute the same object of $D(R)$. Observe that for $i_2 \geq a - 1$ and $n < 2a - 1$ the map
\[ S^{n + 1, i_2 + 1, i_2 + 1} \to S^{n, i_2 + 1, i_2 + 1} \]
is a quasi-isomorphism because left and right hand side are quasi-isomorphic to $\text{Tot}(F^{i_2 + 1}K^\bullet \otimes L^\bullet )$. Thus using the short exact sequence of complexes in the previous paragraph we find that $S^{n, a - 1, a - 1} \to S^{n, a - 1, t}$ is a quasi-isomorphism for all $t \geq a - 1$. On the other hand, there are similarly short exact sequences
\[ 0 \to S^{n + 1, i_1, i_1} \to S^{n, i_1 - 1, i_1 - 1} \oplus S^{n, i_1, i_2} \to S^{n, i_1 - 1, i_2} \to 0 \]
of complexes. Similarly to the above, we note that for $i_1 \leq a - 1$ the map
\[ S^{n + 1, i_1, i_1} \to S^{n, i_1 - 1, i_1 - 1} \]
is a quasi-isomorphism because left and right hand side are quasi-isomorphic to $\text{Tot}(K^\bullet \otimes F^{n - i_1 + 1}L^\bullet )$. Thus we conclude that $S^{n, s, t} \to S^{n, a - 1, t}$ is a quasi-isomorphism for all $s \leq a - 1$. Combined we find that $S^{n, s, t}$ for $s \leq a - 1$ and $t \geq a - 1$ maps quasi-isomorphically to $\text{Tot}(K^\bullet \otimes _ R L^\bullet )$. This finishes the proof of the lemma.
$\square$
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