= Solution
Take $N=N_G(D)$ and $b=\operatorname{Br}_D(e)$. If $eM=M$, the <Nagao module theorem> with $K=N$ puts every vertex-$D$ summand of $M\downarrow_N$ inside $bM$. In particular its <Green correspondence> summand $M'$ satisfies $bM'=M'$.
Conversely a <central idempotent> acts as zero or one on the <indecomposable module> $M$. If $eM=0$, apply the same theorem to $1-e$. Its Brauer image is $1-b$, so $(1-b)M'=M'$, incompatible with $bM'=M'$. Thus
$$
\boxed{eM=M\quad\Longleftrightarrow\quad\operatorname{Br}_D(e)M'=M'.}
$$
For the unheaded induction request, write $\mathcal X=\{D\cap{}^gD:g\notin N\}$. We prove the sharper intersection bound, not just a drop in <module vertex> size. We use <Green correspondence> on permutation bimodules and the usual <module vertex> test for a permutation-module summand: its <Brauer quotient of a module> at $P$ is nonzero precisely when a <module vertex> contains a conjugate of $P$. This test follows by restricting to the $p$-group $P$, decomposing the permutation basis into its orbits, and noting that proper-orbit sums are precisely the proper-subgroup transfer terms.
Consider $kG$ as a $k[G\times D]$-module, with $(g,d)a=gad^{-1}$. It is the transitive <permutation module> induced from $\Delta D$. Set $L=G\times D$ and $L_0=N\times D$; the latter contains $N_L(\Delta D)$. Its restriction decomposes as
$$
kG\downarrow_{L_0}=kN\oplus\bigoplus_{g\in N\backslash G/D,\ g\notin N}k[NgD].
$$
Each outside term has stabilizer a conjugate of a <subgroup> of $\Delta D$. If a diagonal <subgroup> $\Delta Q\le\Delta D$ embeds into one of these stabilizers, then $Q\le D\cap{}^gD$ for an outside representative. Thus its diagonal <module vertices> are in the intersection error family.
Every indecomposable $L_0$-summand of $kN$ has <module vertex> $\Delta D$. To justify this, its endomorphisms are right multiplications by elements of $(kN)^D$. The coefficient projection $\operatorname{Br}_D^N$ on this <algebra> has nilpotent kernel: every nonsingleton $D$-orbit sum lies in the kernel of $kN\to k(N/D)$, by part 1(b), and that kernel is nilpotent by part 1(a). Hence no nonzero <idempotent> endomorphism can have zero Brauer image. Its image summand has nonzero <Brauer quotient of a module> at $\Delta D$, and its <module vertex>, already contained in a conjugate of $\Delta D$, must be $\Delta D$.
Apply the general <Green correspondence> from part (a) to $L,L_0$. If an indecomposable summand of $kG$ has <module vertex> outside the intersection error family, its correspondent must be a summand of $kN$, since the outside double-coset <modules> have only error-family <module vertices>. It therefore has <module vertex> $\Delta D$. Consequently \b[every summand of $kG$ not having <module vertex> $\Delta D$ has <module vertex> conjugate into an outside diagonal intersection]. An intersection
$$
\Delta D\cap{}^{(g,d)}\Delta D
$$
with $(g,d)\notin L_0$ projects in its first coordinate into $D\cap{}^gD$ with $g\notin N$; this is the error family required in the question.
Now form the split permutation bimodule
$$
W=(1-e)kG b.
$$
Its <Brauer quotient of a module> at $\Delta D$ is zero. Indeed $kG(\Delta D)=kC_G(D)$, left multiplication by $e$ becomes multiplication by $\operatorname{Br}_D(e)=b$, and right multiplication by $b$ remains multiplication by $b$. Therefore
$$
W(\Delta D)=(1-b)kC_G(D)b=0.
$$
Here $b$ is central in $kC_G(D)$, so the last product is zero. Every indecomposable bimodule summand of $W$ consequently has <module vertex> in the outside intersection family, by the preceding paragraph, and has trivial <module source> since it is a permutation-module summand.
Since $V$ has vertex $D$, necessarily $D\le K$. Let $S$ be a <module source> of $V$ at $D$. The splitting $V\mid kK\otimes_{kD}S$ remains a splitting after applying $b$, since $bV=V$. Inducing to $G$ and applying $1-e$ gives
$$
(1-e)\operatorname{Ind}_K^G V\mid W\otimes_{kD}S.
$$
A trivial-source bimodule summand with <module vertex> conjugate into $\Delta Q$ is a summand of $\operatorname{Ind}_{\Delta Q}^{G\times D}k$. Tensoring with $S$ turns this into a <module> induced to $G$ from $Q$ (with the appropriate conjugate restriction of $S$). Thus every summand on the right is relatively projective for some $Q\le D\cap{}^gD$, $g\notin N$. Each $V_j$ lies in the left-hand side, and <Krull–Schmidt theorem> now gives
$$
\boxed{D_j\le_G D\cap{}^gD\quad\text{for some }g\in G\setminus N_G(D).}
$$
This proves <Juhász induction refinement>. If there are no outside elements, the error family is empty and the complementary <module> is zero.
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