Proof.
Choose a functorial injective embeddings $i_ M : M \to I(M)$, see Theorem 19.11.7. For every complex $M^\bullet $ denote $J^\bullet (M^\bullet )$ the complex with terms $J^ n(M^\bullet ) = I(M^ n) \oplus I(M^{n + 1})$ and differential
\[ d_{J^\bullet (M^\bullet )} = \left( \begin{matrix} 0
& 1
\\ 0
& 0
\end{matrix} \right) \]
There exists a canonical injective map of complexes $u_{M^\bullet } : M^\bullet \to J^\bullet (M^\bullet )$ by mapping $M^ n$ to $I(M^ n) \oplus I(M^{n + 1})$ via the maps $i_{M^ n} : M^ n \to I(M^ n)$ and $i_{M^{n + 1}} \circ d : M^ n \to M^{n + 1} \to I(M^{n + 1})$. Hence a short exact sequence of complexes
\[ 0 \to M^\bullet \xrightarrow {u_{M^\bullet }} J^\bullet (M^\bullet ) \xrightarrow {v_{M^\bullet }} Q^\bullet (M^\bullet ) \to 0 \]
functorial in $M^\bullet $. Set
\[ \mathbf{N}^\bullet (M^\bullet ) = C(v_{M^\bullet })^\bullet [-1]. \]
Note that
\[ \mathbf{N}^ n(M^\bullet ) = Q^{n - 1}(M^\bullet ) \oplus J^ n(M^\bullet ) \]
with differential
\[ \left( \begin{matrix} - d^{n - 1}_{Q^\bullet (M^\bullet )}
& - v^ n_{M^\bullet }
\\ 0
& d^ n_{J^\bullet (M)}
\end{matrix} \right) \]
Hence we see that there is a map of complexes $j_{M^\bullet } : M^\bullet \to \mathbf{N}^\bullet (M^\bullet )$ induced by $u$. It is injective and factors through an injective subobject by construction. The map $j_{M^\bullet }$ is a quasi-isomorphism as one can prove by looking at the long exact sequence of cohomology associated to the short exact sequences of complexes above.
$\square$
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