[step:Two-case algebraic split (borderline) and conclusion]
Fix $s > 0$. We need an algebraic input that converts the logarithmic sum ($\dagger'$) into a polynomial $s$-content sum. The function
\begin{align*}
g(r) := r^s \cdot (\log(R_0' / r))^{n - 1}, \qquad r \in (0, R_0'/e],
\end{align*}
is bounded. Indeed, set $u := \log(R_0' / r)$, so $r = R_0' e^{-u}$ and $u \in [1, \infty)$ corresponds to $r \in (0, R_0'/e]$. Then $g = (R_0')^s e^{-s u} u^{n - 1}$, a continuous function of $u \in [1, \infty)$ that equals $(R_0'/e)^s$ at $u = 1$ and tends to $0$ as $u \to \infty$ (since $e^{-su}$ decays faster than $u^{n-1}$ grows, for any $s > 0$). Hence $g$ attains its supremum on $[1, \infty)$ and
\begin{align*}
M := M(n, s, R_0') := \sup_{r \in (0, R_0'/e]} r^s (\log(R_0' / r))^{n - 1} < \infty.
\end{align*}
For every $r \in (0, R_0'/e]$,
\begin{align*}
r^s \le M \cdot (\log(R_0' / r))^{-(n - 1)}. \tag{$\flat$}
\end{align*}
Since $r_k \le R_0'/e$ for every $k$ in the cover from Step 5, ($\flat$) applies uniformly across $k$.
Two-case split at scale $\delta$. We do not refine the cover; the borderline case fails the same naive-refinement test as the subcritical case (the volume-comparison factor $(r_k/\rho)^n$ is not absorbed by the logarithmic ratio $(\log(R_0'/r_k))^{n-1} / (\log(R_0'/\rho))^{n-1}$ into a dimension-only constant). Partition
\begin{align*}
K_{\text{small}} := \{k : r_k \le \delta/2\}, \qquad K_{\text{large}} := \{k : r_k > \delta/2\}.
\end{align*}
For $k \in K_{\text{small}}$, keep the ball ($\operatorname{diam} = 2 r_k \le \delta$), contributing $(2 r_k)^s$. For $k \in K_{\text{large}}$, cover by $N_k \le N_{\text{pack}}(n) (2 r_k / \delta)^n$ balls of radius $\delta/2$ (each diameter $\delta$, admissible), contributing $N_k \delta^s$. Hence
\begin{align*}
\mathcal{H}^s_\delta(E) \le 2^s \sum_{k \in K_{\text{small}}} r_k^s + N_{\text{pack}}(n)\, 2^n\, \delta^{s - n} \sum_{k \in K_{\text{large}}} r_k^n.
\end{align*}
Small balls. By ($\flat$) and ($\dagger'$),
\begin{align*}
\sum_{k \in K_{\text{small}}} r_k^s \le M \sum_{k \in K_{\text{small}}} (\log(R_0'/r_k))^{-(n-1)} \le M \cdot C(n)\, \eta.
\end{align*}
Large balls. For $k \in K_{\text{large}}$ we have $\delta/2 < r_k \le R_0'/e$, in particular the radii lie in a bounded interval. The map $r \mapsto r^n \cdot (\log(R_0'/r))^{n-1}$ is continuous and strictly positive on the closed interval $[\delta/2, R_0'/e]$, hence attains its supremum; let
\begin{align*}
M_n := M_n(n, \delta, R_0') := \sup_{r \in [\delta/2, R_0'/e]} r^n (\log(R_0'/r))^{n - 1} < \infty.
\end{align*}
(For $\delta/2 \ge R_0'/e$, the set $K_{\text{large}}$ is empty and this contribution is zero; we may assume $\delta/2 < R_0'/e$, taking $\delta$ sufficiently small.) Then for $k \in K_{\text{large}}$,
\begin{align*}
r_k^n \le M_n \cdot (\log(R_0'/r_k))^{-(n-1)},
\end{align*}
and summing,
\begin{align*}
\sum_{k \in K_{\text{large}}} r_k^n \le M_n \sum_{k \in K_{\text{large}}} (\log(R_0'/r_k))^{-(n-1)} \le M_n \cdot C(n)\, \eta.
\end{align*}
Conclusion. Combining,
\begin{align*}
\mathcal{H}^s_\delta(E) \le \big[ 2^s M + N_{\text{pack}}(n)\, 2^n\, \delta^{s - n}\, M_n \big] \cdot C(n)\, \eta =: C(n, s, \delta, R_0')\, \eta.
\end{align*}
The bracketed factor depends on $n, s, \delta, R_0'$ but not on the cover and not on $\varphi$. Letting $\eta \to 0^+$ at fixed $n, s, \delta, R_0'$ gives $\mathcal{H}^s_\delta(E) = 0$. This holds for every $\delta > 0$, so $\mathcal{H}^s(E) = 0$. Since $s > 0$ was arbitrary, $\dim_{\mathcal{H}}(E) = 0$, for bounded $E$. The countable union argument from Step 1 extends the conclusion to arbitrary Borel $E$.
This completes the proof in both ranges.
[/step]