Is the zig-zag lemma natural if the diagram of short exact sequences is only commutative up to homotopy?

Consider two exact sequences of chain complexes (of abelian groups) with morphisms of chain complexes $f, g, h$ between them:$\require{AMScd}$ $$ \begin{CD} 0 @>>> A_\bullet @>\alpha>> B_\bullet @>\beta>> C_\bullet @>>> 0\\ & @VfVV @VgVV @VhVV \\ 0 @>>> X_\bullet @>\xi>> Y_\bullet @>\nu>> Z_\bullet @>>> 0 \label{d1} \tag{1} \end{CD} $$ If both squares in the diagram are commutative, the naturality of the zig-zag lemma gives us a morphism between the two long exact sequences, i. e. the diagram $$ \begin{CD} @>>> H_n(A_\bullet) @>>> H_n(B_\bullet) @>>> H_n(C_\bullet) @>\delta>> H_{n-1}(A_\bullet) @>>> \\ &@Vf_*VV @Vg_*VV @Vh_*VV @Vf_*VV \\ @>>> H_n(X_\bullet) @>>> H_n(Y_\bullet) @>>> H_n(Z_\bullet) @>\delta>> H_{n-1}(X_\bullet) @>>> \end{CD} $$ is commutative. Actually, the only nontrivial part of this is the commutativity of the the right square with the connecting morphisms $\delta$.

Question: What happens if the squares in diagram (\ref{d1}) are only commutative up to homotopy? My intuition told me that the zig-zag lemma should still be true, but I can't manage to prove it and I am actually starting to doubt whether this could really work. Again, the only nontrivial part is the commutativity of the squares with the $\delta$s.

Edit: I guess that this is equivalent to asking whether the zig-zag lemma is still valid in the homotopy category of chain complexes

This answer gives the following counterexample. $\newcommand\toby\xrightarrow$

Let $k$ be a field. Take $B= \cdots\to 0 \to k \toby{1} k \to 0 \to \cdots$, with $k$s in degrees $0$ and $1$. (Arrows going to the right for notational convenience, but still with homological grading.) Let $A$ be the subcomplex $\cdots \to 0 \to 0 \to k \to 0 \to \cdots$, with a $k$ in degree $0$. Then $C=B/A$ has a $k$ in degree $1$. Then the diagram $$ \require{AMScd} \begin{CD} 0 @>>> A @>>> B @>>> B/A @>>> 0 \\ @. @VV1V @VV1V @VV0V @. \\ 0 @>>> A @>>> B @>>> B/A @>>> 0 \\ \end{CD} $$ commutes up to homotopy. The first square clearly commutes, and $B$ is contractible, the nullhomotopy being $\cdots \leftarrow 0 \leftarrow k \overset{1}{\leftarrow} k \leftarrow 0 \leftarrow \cdots$, so any two maps out of $B$ are homotopic.

Then the LES in homology is the following $$ \begin{CD} 0=H_1(B) @>>> H_1(B/A) @>\delta>> H_0(A) @>>> H_0(B)= 0\\ @. @VV0V @VV1V @. \\ 0=H_1(B) @>>> H_1(B/A) @>\delta>> H_0(A) @>>> H_0(B) = 0,\\ \end{CD} $$ which becomes $$ \begin{CD} 0 @>>> k @>1>> k @>>> 0\\ @. @VV0V @VV1V @. \\ 0 @>>> k @>1>> k @>>> 0,\\ \end{CD} $$ which does not commute.