Contradictory statements in a proof on m-eternal domination Given the upper bound $\gamma^{\infty}_{m}(G) \leq 2\gamma(G)$ for the m-eternal domination number, I'm studying a proof which shows that for every possible value of $\gamma \in \mathbb{N}^{+}$ there exists a graph G s.t. $\gamma(G) = \gamma$ and $\gamma^{\infty}_{m}(G) = 2\gamma$ (the m-et. dom. number reaches its bound).
The proof goes by induction on $\gamma$, but there are some contradictory statements to me which I truly can't understand.
For $\gamma = 2$, consider $C_{6}$: let u and v be two disjoint vertices in $C_{6}$ at distance three apart, I picked $v_{3}$ and $v_{6}$ for example. We add two new internally disjoint u-v paths of length three to form the graph G. I assumed that something like this should be good (you can see both $C_{6}$ and G in the picture: $\{v_{3},v_{5},v_{2},v_{6}\}$ and $\{v_{3},v_{1},v_{4},v_{6}\}$ are the two new paths that I added).

In G $\{u,v\}$ ($\{v_{3},v_{6})$ in my example) is a dominating set, so $\gamma(G) = 2$. It remains to show that $\gamma^{\infty}_{m}(G) = 4$.
Let D be a dominating set of G s.t. $|D| = 3$. Suppose $v_{3} \notin D$: the proof tries to reach a contradiction to show that both $v_{3}$ and $v_{6}$ must be in D. It's said that the set of $v_{3}$'s open neighborhoods N(u) is an independent set of order four, but how is this possible? If we have to add two internally disjoint paths from $v_{3}$ and $v_{6}$ and $|N(u)|$ must be 4 at the same time it's necessary to add the edges $v_{3}v_{5}$ and $v_{3}v_{1}$ and there's no other possible choice, I think. How can we assume that N(u) is an independent set of nodes that have u as only common neighborhood? Is there something I did wrong?
The actual proof is from paper https://arxiv.org/pdf/1407.5228.pdf, proposition 12
 A: You're supposed to add two paths with entirely new vertices, getting a graph like the one below:

Here, $N(u)$ is the independent set $\{1,3,5,7\}$ and $N(v)$ is the independent set $\{2,4,6,8\}$.
A set $D \subseteq V$ with $|D|=3$ and $u \notin D$ cannot dominate all four of $\{1,3,5,7\}$; we'd need at least one vertex from the interior of each of the four paths. Similarly, a set $D\subseteq V$ with $|D|=3$ and $v\notin D$ cannot dominate all four of $\{2,4,6,8\}$.
All sets $D\subseteq V$ with $|D|=3$ and $u,v \in D$ are symmetric to the set $\{u,v,1\}$ under an automorphism of the graph, so to finish the proof, it's enough to show that $\{u,v,1\}$ is not m-eternal dominating. Well, suppose vertex $4$ is attacked. Then the guard on $v$ must move to $4$, leaving $6$ and $8$ unprotected; the guard on $u$ can move to $5$ or to $7$ to protect one of them, but the guard on $1$ can't do anything useful.
Similarly, here's a diagram of the $\gamma=4$ case, after which you'll start to get the general idea of the construction:

Unfortunately, it is not clear to me what argument is glossed over by the phrase "It can be shown similar to the previous case" in the proof. In particular, it is no longer true that any dominating set of size less than $2\gamma$ has to contain the minimum dominating set $\{u_0, u_3, u_6, u_9\}$; for example, we could replace $u_0$ by its neighborhood $\{u_1, u_{11}, v_1, v_{11}\}$ and still be under the limit.
The proof becomes easier if we add many length-three paths between every $u_{3i}$ and $u_{3i+3}$ (and between $u_{3\gamma-3}$ and $u_0$), not just one. In particular, if you add $\gamma-1$ paths, then the neighborhood of each $u_{3i}$ is an independent set of $2\gamma$ vertices, and so certainly all of $\{u_0, u_3, u_6, \dots, u_{3\gamma-3}\}$ need to have guards on them. If there are fewer than $2\gamma$ guards, then one of these guards does not have backup: some guard on $u_{3i}$ does not have a guard on any adjacent vertex. Attack any vertex protected exclusively by that guard, and vertex $u_{3i}$ becomes empty, in which case not all of $N(u_{3i})$ can be protected.
