If I am not mistaken, then the following recursive construction disproves statement A:

- Let $G_1$ consist of a $K_8$ and an additional vertex $r_1$ connected to half of the vertices of this $K_8$
- For $k > 1$ take $2^{2^k}$ copies of $G_{k-1}$, connect each pair of copies of $r_{k-1}$ by an edge, and add a vertex $r_k$ which is incident to half of the copies of $r_{k-1}$

(connecting $r_k$ only to half of the copies has the sole purpose of making the numbers in the induction below a bit prettier)

**Claim 1:** $G_k$ has $\frac{k+3}{2} 2^{2^{k+1}}$ edges and $2\cdot 2^{2^{k+1}}$ of these edges are contained in copies of $G_1$.

We use induction on $k$. For $k=1$ note that $G_1$ has ${8 \choose 2} + 4 = 32 = \frac{1+3}{2} 2^{2^{1+1}}$ edges and all of them are contained in a copy of $G_1$.

For the induction step, note that the $2^{2^k}$ copies of $r_{k-1}$ form a complete graph. Together with the $\frac{2^{2^k}}{2}$ edges incident to $r_k$ the number of edges of $G_k$ which are not contained in any copy of $G_{k-1}$ is thus $${2^{2^k}\choose 2} + \frac{2^{2^k}}{2} = \frac 12 (2^{2^k})^2 = \frac{2^{2^{k+1}}}{2}.$$
Since $G_k$ contains $2^{2^k}$ copies of $G_{k-1}$ each of which by induction hypothesis contains $\frac{k+2}{2} 2^{2^{k}}$ edges, the total number of edges in $G_k$ is
$$
\frac{2^{2^{k+1}}}{2} + 2^{2^k} \frac{k+2}{2} 2^{2^{k}} = \frac{k+3}{2} 2^{2^{k+1}}
$$
as claimed. For the number of edges contained in a copy of $G_1$ simply note that all of them are contained in one of the $2^{2^k}$ copies of $G_{k-1}$ each of which contains $2\cdot 2^{2^k}$ such edges, giving
$$
2^{2^k} \cdot 2 \cdot 2^{2^k} = 2\cdot 2^{2^{k+1}}
$$
such edges in total. This finishes the proof of Claim 1.

**Claim 2:** Let $S$ be a connected set of vertices of $G_k$ with maximal edge boundary.
Then $V(G_k) \setminus S$ only consists of vertices in copies of $G_1$ and $r_k$, and $r_1$ is only in $V(G_k) \setminus S$ if $k=1$. This immediately implies that
$$
|\partial_{\text{edge}} S| \leq 2\cdot 2^{2^{k+1}} + \frac{2^{2^k}}{2}.
$$

To show this, we first note that $G_k$ contains a set $S'$ with $|\partial_{\text{edge}} S| > 2^{2^{k+1}}$ because in every copy of $G_1$ we can pick a connected set containing the respective copy of $r_1$ whose edge boundary contains $19$ of the $32$ edges and then add all vertices not contained in any copy of $G_1$. In particular, the set $S$ whose edge boundary is maximal must contain at least one of the copies of $r_{k-1}$, otherwise $S$ would be contained in some copy of $G_k$ and thus by induction its edge boundary would contain at most $2\cdot 2^{2^{k}} + 2^{2^{k-1}} < 2^{2^{k+1}}$ edges.

Now assume that some copy of $r_{k-1}$ is not contained in $S$. This copy of $r_{k-1}$ has at most $2^{2^k}$ neighbours in $S$. Thus adding this copy and a connected subset of the corresponding copy of $G_{k-1}$ whose edge boundary contains strictly more than $2^{2^{k}}$ edges to $S$ increases the edge boundary while the set stays connected, thereby contradicting maximality of $S$. We have thus shown that $S$ contains all copies of $r_{k-1}$, and by iterating this argument we see that $S$ must contain all copies of $r_{j}$ for every $j<k$.