[proofplan]
We use the standard unaugmented cover double complex. Its vertical-first spectral sequence has first page given by the homology of the intersections $K_\sigma$. The connectivity hypotheses force all terms with positive vertical degree and total degree at most $r$ to vanish; the surviving degree-zero row is exactly the simplicial chain complex of the nerve. The horizontal-first spectral sequence computes the homology of the union $K$, because each simplex of $K$ contributes the chain complex of a full simplex on the cover elements containing it. Comparing the two spectral sequences gives the low-degree isomorphisms and the edge surjection in degree $r+1$; reduced degree zero follows by taking kernels of the augmentation maps.
[/proofplan]
[step:Construct the cover double complex]
Choose a total ordering of the finite set $A$, and write $\mathcal U:=\{K_a\}_{a\in A}$. The nerve $N(\mathcal U)$ has as simplices the nonempty finite subsets $\sigma\subset A$ such that
\begin{align*}
K_\sigma:=\bigcap_{a\in\sigma}K_a
\end{align*}
is nonempty. For $p,q\geq 0$, define
\begin{align*}
D_{p,q}:=\bigoplus_{\sigma\in N(\mathcal U),\ |\sigma|=p+1} C_q(K_\sigma;k),
\end{align*}
and set $D_{p,q}=0$ otherwise. The vertical differential is the direct sum of the simplicial boundary maps inside the complexes $K_\sigma$. The horizontal differential is the alternating Cech differential: on the summand indexed by $\sigma=\{a_0<\cdots<a_p\}$ it is
\begin{align*}
d_h=\sum_{m=0}^{p}(-1)^m\iota_{\sigma,\sigma\setminus\{a_m\}},
\end{align*}
where
\begin{align*}
\iota_{\sigma,\tau}:C_q(K_\sigma;k)\to C_q(K_\tau;k)
\end{align*}
is induced by the inclusion $K_\sigma\subset K_\tau$.
The vertical and horizontal differentials commute because inclusions of subcomplexes commute with simplicial boundary maps. Define
\begin{align*}
\operatorname{Tot}_n(D):=\bigoplus_{p+q=n}D_{p,q}
\end{align*}
with total differential $d=d_h+(-1)^p d_v$ on $D_{p,q}$. Then $d^2=0$.
[guided]
The construction uses two indices. The cover index is recorded by a simplex $\sigma\subset A$ of the nerve, and the chain index is recorded inside the intersection
\begin{align*}
K_\sigma:=\bigcap_{a\in\sigma}K_a.
\end{align*}
For $p,q\geq 0$ the double complex group is
\begin{align*}
D_{p,q}:=\bigoplus_{\sigma\in N(\mathcal U),\ |\sigma|=p+1} C_q(K_\sigma;k).
\end{align*}
The vertical differential is the usual simplicial boundary on each $C_q(K_\sigma;k)$. The horizontal differential is the Cech boundary: for $\sigma=\{a_0<\cdots<a_p\}$,
\begin{align*}
d_h=\sum_{m=0}^{p}(-1)^m\iota_{\sigma,\sigma\setminus\{a_m\}},
\end{align*}
where
\begin{align*}
\iota_{\sigma,\tau}:C_q(K_\sigma;k)\to C_q(K_\tau;k)
\end{align*}
comes from $K_\sigma\subset K_\tau$. The total complex is
\begin{align*}
\operatorname{Tot}_n(D):=\bigoplus_{p+q=n}D_{p,q},
\end{align*}
with $d=d_h+(-1)^p d_v$. Since $d_vd_h=d_hd_v$, this total differential satisfies $d^2=0$.
[/guided]
[/step]
[step:Compute the vertical-first spectral sequence]
Filter $\operatorname{Tot}(D)$ by horizontal degree. Since $A$ is finite, this filtration is finite in each total degree, so the associated spectral sequence converges to $H_*(\operatorname{Tot}(D))$. Its first page is
\begin{align*}
E_1(p,q)\cong \bigoplus_{\sigma\in N(\mathcal U),\ |\sigma|=p+1} H_q(K_\sigma;k).
\end{align*}
For $q=0$, the hypothesis with $q=0$ says that $K_\sigma$ is connected whenever $p\leq r+1$, because then
\begin{align*}
0\leq r-p+1=r-|\sigma|+2.
\end{align*}
Thus, for $0\leq p\leq r+1$, the $q=0$ row agrees with the ordinary simplicial chain complex $C_p(N(\mathcal U);k)$.
For $q>0$ and $p+q\leq r$, the same inequality gives
\begin{align*}
q\leq r-p<r-p+1=r-|\sigma|+2,
\end{align*}
so the homological connectivity hypothesis gives $H_q(K_\sigma;k)=0$. Therefore
\begin{align*}
E_1(p,q)=0
\end{align*}
for every $q>0$ with $p+q\leq r$.
It follows that, in total degrees at most $r$, the only possible nonzero terms lie on the nerve row $q=0$. Hence
\begin{align*}
E_2(p,0)\cong H_p(N(\mathcal U);k)
\end{align*}
for $p\leq r+1$, and the terms with $p\leq r$ survive to $E_\infty$. In total degree $r+1$, no differential can enter $E_s(r+1,0)$ because there are no negative vertical degrees, and every differential leaving it lands in total degree $r$ with positive vertical degree, where the terms vanish. Hence
\begin{align*}
E_\infty(r+1,0)\cong E_2(r+1,0)\cong H_{r+1}(N(\mathcal U);k).
\end{align*}
[guided]
Taking vertical homology first gives
\begin{align*}
E_1(p,q)\cong \bigoplus_{\sigma\in N(\mathcal U),\ |\sigma|=p+1} H_q(K_\sigma;k).
\end{align*}
When $q=0$ and $p\leq r+1$, we have
\begin{align*}
0\leq r-p+1=r-|\sigma|+2,
\end{align*}
so the hypothesis says each $K_\sigma$ is connected. Thus the $q=0$ row is the nerve chain complex $C_p(N(\mathcal U);k)$. When $q>0$ and $p+q\leq r$, we have
\begin{align*}
q\leq r-p<r-p+1=r-|\sigma|+2,
\end{align*}
and therefore
\begin{align*}
E_1(p,q)=0.
\end{align*}
Consequently the low-degree part of the spectral sequence has only the row $q=0$. This gives
\begin{align*}
E_2(p,0)\cong H_p(N(\mathcal U);k)
\end{align*}
for $p\leq r+1$, and no later differential changes the groups in total degree at most $r+1$. In the next degree the edge term is still
\begin{align*}
E_\infty(r+1,0)\cong H_{r+1}(N(\mathcal U);k),
\end{align*}
because differentials leaving it would land in vanished positive rows of total degree $r$.
[/guided]
[/step]
[step:Compute the horizontal-first spectral sequence]
Now take horizontal homology first. Fix $q\geq 0$ and a $q$-simplex $\tau$ of $K$. Let
\begin{align*}
A(\tau):=\{a\in A: \tau\subset K_a\}.
\end{align*}
This set is nonempty because the subcomplexes $K_a$ cover $K$. The copies of the basis vector $\tau$ in the row $D_{*,q}$ are indexed by the nonempty subsets $\sigma\subset A(\tau)$, and the horizontal differential is the simplicial boundary of the full simplex on the ordered vertex set $A(\tau)$.
A nonempty full simplex has homology $k$ in degree zero and zero homology in all positive degrees. Therefore the horizontal-first spectral sequence has
\begin{align*}
{}'E_1(0,q)\cong C_q(K;k)
\end{align*}
and
\begin{align*}
{}'E_1(p,q)=0
\end{align*}
for every $p>0$. Under this identification, the remaining vertical differential is the ordinary simplicial boundary of $K$: if $\partial_j\tau$ is a face of $\tau$, then $A(\tau)\subset A(\partial_j\tau)$, so the augmentation class for the full simplex on $A(\tau)$ maps to the augmentation class for the full simplex on $A(\partial_j\tau)$ with the usual simplicial sign.
Thus the horizontal-first spectral sequence collapses to the simplicial chain complex of $K$, and
\begin{align*}
H_n(\operatorname{Tot}(D))\cong H_n(K;k)
\end{align*}
for every $n\geq 0$.
[guided]
Fix a $q$-simplex $\tau$ of $K$ and define
\begin{align*}
A(\tau):=\{a\in A: \tau\subset K_a\}.
\end{align*}
Because the cover is by subcomplexes, $A(\tau)$ is nonempty. The copies of $\tau$ in
\begin{align*}
D_{p,q}=\bigoplus_{\sigma\in N(\mathcal U),\ |\sigma|=p+1} C_q(K_\sigma;k)
\end{align*}
are exactly indexed by the nonempty $\sigma\subset A(\tau)$. Horizontally, these copies form the chain complex of the full simplex on $A(\tau)$. Hence the horizontal homology contributes one copy of $k$ in degree $p=0$ and no copies in degrees $p>0$:
\begin{align*}
{}'E_1(0,q)\cong C_q(K;k)
\end{align*}
and
\begin{align*}
{}'E_1(p,q)=0.
\end{align*}
The surviving vertical differential is the usual boundary, since $A(\tau)\subset A(\partial_j\tau)$ for every face $\partial_j\tau$. Therefore
\begin{align*}
H_n(\operatorname{Tot}(D))\cong H_n(K;k)
\end{align*}
for every $n\geq 0$.
[/guided]
[/step]
[step:Compare the two computations]
The vertical-first computation gives a filtration on $H_n(\operatorname{Tot}(D))$ whose associated graded pieces are the terms $E_\infty(p,q)$ with $p+q=n$. For $n\leq r$, all pieces with $q>0$ vanish, so the only nonzero piece is
\begin{align*}
E_\infty(n,0)\cong H_n(N(\mathcal U);k).
\end{align*}
Combining this with the horizontal-first computation gives natural isomorphisms
\begin{align*}
H_n(K;k)\cong H_n(N(\mathcal U);k)
\end{align*}
for $0\leq n\leq r$.
For $n=r+1$, the edge projection from a filtered [vector space](/page/Vector%20Space) to its associated graded quotient gives a natural surjection
\begin{align*}
H_{r+1}(\operatorname{Tot}(D))\to E_\infty(r+1,0).
\end{align*}
Using the identifications above, this is a natural surjection
\begin{align*}
H_{r+1}(K;k)\to H_{r+1}(N(\mathcal U);k).
\end{align*}
Finally pass from ordinary homology to reduced homology. In positive degrees there is no change. In degree zero, the ordinary $H_0$ isomorphism identifies the number of connected components of $K$ with the number of connected components of $N(\mathcal U)$, and the induced map preserves the augmentation to $k$ on each side. Taking kernels of the augmentation maps therefore gives
\begin{align*}
\widetilde H_0(K;k)\cong \widetilde H_0(N(\mathcal U);k).
\end{align*}
Together with the positive-degree statements, this proves the asserted reduced homology isomorphisms for $i\leq r$ and the asserted surjection in degree $r+1$.
[guided]
For $n\leq r$, the vertical-first spectral sequence has no positive-row pieces, so
\begin{align*}
E_\infty(n,0)\cong H_n(N(\mathcal U);k).
\end{align*}
The horizontal-first computation gives
\begin{align*}
H_n(\operatorname{Tot}(D))\cong H_n(K;k),
\end{align*}
so we get
\begin{align*}
H_n(K;k)\cong H_n(N(\mathcal U);k)
\end{align*}
for $0\leq n\leq r$. In degree $r+1$, the edge quotient gives
\begin{align*}
H_{r+1}(\operatorname{Tot}(D))\to E_\infty(r+1,0),
\end{align*}
and this becomes
\begin{align*}
H_{r+1}(K;k)\to H_{r+1}(N(\mathcal U);k).
\end{align*}
It is surjective because it is the projection to an associated graded quotient. For reduced homology, positive degrees are unchanged and degree zero is obtained by taking kernels of the augmentation maps, giving
\begin{align*}
\widetilde H_0(K;k)\cong \widetilde H_0(N(\mathcal U);k).
\end{align*}
[/guided]
[/step]
