Every block contains a module whose vertex is a full defect group

Take $b$ from F1. It is nonzero, so a proper left ideal of largest dimension in $b$ gives a nonzero simple quotient $S$. Extending by zero on the other blocks makes it a simple $kN$-module. By F6, $D=O_p(N)$, in particular $D⊴N$, and F5 makes $D$ act trivially on $S$. Thus $S$ is a simple module of the finite-dimensional quotient group algebra $k[N\mathrm{/}D]$.

Take its projective cover $P\twoheadrightarrow S$ over $k[N\mathrm{/}D]$ using F3; it is finite dimensional, nonzero and indecomposable by F4. Inflate to $N$. The action factors through the quotient, so its submodules and endomorphisms are unchanged and it stays indecomposable. The central idempotents in F9 decompose $P$ into block pieces. The piece for $b$ maps onto $S$, since the idempotent of $b$ acts as identity on $S$. It is therefore nonzero, and indecomposability forces that piece to be all of $P$. Thus $P$ lies in $b$.

F3 realizes $P$ as a summand of a finite free $k[N\mathrm{/}D]$-module. Inflating $k[N\mathrm{/}D]$ gives the permutation module ${\mathrm{Ind}}_D^{N}k$: the map $n\otimes 1\mapsto nD$ identifies their bases and actions. Hence $P$ is relatively $D$-projective. By F11 and F12 it has a vertex $R\le D$, after conjugating in $N$ (normality of $D$ retains this containment). On restriction to $D$, $P$ is a nonzero direct sum of trivial one-dimensional modules, since all of $D$ acts trivially. The trivial $kD$-module has vertex $D$: for $E<D$, every scalar endomorphism has relative trace $[D:E]a=0$ in characteristic $p$, whereas the identity is nonzero, so F10 excludes relative $E$-projectivity. F8 applied to $R\le D$ says the restriction of $P$ has a summand with vertex $R$. All its indecomposable summands are trivial and have vertex $D$, forcing $R=D$.

Apply F7 under A1 to $P$ for $N=N_G(D)$. Its inverse correspondent $V$ is a nonzero indecomposable $kG$-module with vertex $D$, and $P\mid {\mathrm{Res}}_N^{G}V$. F2 applies with $Q=D$, since $D{C}_G(D)\le N$, and identifies the block of $V$ as $b^G=B$. This is the desired module. If $D=1$, the correspondence is identity and the cover already provides the module in $B$. The nonzero simple quotient guarantees no zero object enters. The finite ideal, cover and trace calculations are choice-free; AC is inherited only from F7's declared finite-length chain and selection argument.