A Cartan-Eilenberg resolution totalizes to an injective replacement

The possible summands of $T^n$ have $b\le p\le n$, hence form a finite biproduct of injectives. For $n<b$ this is zero. The augmentation takes $K^p$ into $I^{p,0}$; $vϵ=0$ and $hϵ=ϵd_K$ give $De=e{d}_K$.

Adjoin $K^p$ in vertical degree $-1$. Set $A^{p,-1}=K^p$ and $A^{p,q}=I^{p,q}$ for $q\ge 0$, with vertical augmentation $ϵ$. All columns of $A$ are exact. The total object $U^n=T^n\oplus K^{n+1}$ with the signed differential is isomorphic to $\mathrm{Cone}(e{)}^n$, whose differential is $(t,k)\mapsto (Dt+ek,-d_{K}k)$. Explicitly send the $K^{n+1}$ summand of $U^n$ to $(-1{)}^{n+1}k$ in the cone; the $T$ summand is unchanged. This verifies both signs, including negative $b$.

Let $F^{m}U$ be the subcomplex consisting of columns $p\ge m$. The quotient $U\mathrm{/}{F}^{m}U$ has only the columns $b,\dots ,m-1$. Its finite descending column filtration has shifted exact columns as successive quotients, hence it is acyclic by repeated application of the long exact sequence. Fix $n$ and take $m>n+2$. Then $F^{m}U$ is zero in degrees $n-1,n,n+1$, since its least total degree is $m-1$. Therefore $H^n(U)=H^n(U\mathrm{/}{F}^{m}U)=0$. This is a finite argument for each degree and requires neither exact filtered colimits nor any infinite limit.

The degreewise split sequence $0\to T\to \mathrm{Cone}(e)\to K[1]\to 0$ has connecting map $H^n(K)\to H^n(T)$ induced by $e$: lift a cycle to the $K$ summand and its cone differential is its image under $e$. The zero cone cohomology in step 3.1 and the long exact sequence thus make every $H^n(e)$ invertible. Finally apply the bounded-below injective theorem with exactly its DC/supplied-extension hypothesis to obtain K-injectivity. The zero complex and one-column case obey the same construction.