[proofplan]
We prove $\mathcal{H}^s(E) \geq \mu(E)/c$ by showing that every admissible cover of $E$ has $s$-dimensional cost at least $\mu(E)/c$. Given any countable cover $\{U_j\}$ of $E$ by sets with $\operatorname{diam}(U_j) \leq \delta < 2r_0$, we enclose each $U_j$ in a ball of radius $\operatorname{diam}(U_j)$, apply the ball condition to bound $\mu(U_j)$, and sum to get $\mu(E) \leq c \sum_j (\operatorname{diam}(U_j))^s$. Taking the infimum over all such covers yields $\mathcal{H}^s_\delta(E) \geq \mu(E)/c$, and the result follows by sending $\delta \to 0$.
[/proofplan]
[step:Enclose each covering set in a ball and apply the mass condition]
Fix $\delta \in (0, 2r_0)$ and let $\{U_j\}_{j=1}^\infty$ be any countable collection of sets covering $E$ with $\operatorname{diam}(U_j) \leq \delta$ for all $j$. We may assume each $U_j$ is non-empty (discarding empty sets does not affect the cover or the sum).
For each $j$, choose a point $x_j \in U_j \cap E$ if $U_j \cap E \neq \varnothing$ (sets with $U_j \cap E = \varnothing$ contribute nothing to covering $E$ and can be discarded). Since $U_j \subset B(x_j, \operatorname{diam}(U_j))$ (every point of $U_j$ is within distance $\operatorname{diam}(U_j)$ of $x_j$), we have
\begin{align*}
\mu(U_j \cap E) \leq \mu(U_j) \leq \mu(B(x_j, \operatorname{diam}(U_j))).
\end{align*}
Since $x_j \in E$ and $\operatorname{diam}(U_j) \leq \delta < 2r_0$, the radius $\operatorname{diam}(U_j)$ satisfies $\operatorname{diam}(U_j) < 2r_0$. For $\operatorname{diam}(U_j) < r_0$, the ball condition gives directly $\mu(B(x_j, \operatorname{diam}(U_j))) \leq c \cdot (\operatorname{diam}(U_j))^s$. For $r_0 \leq \operatorname{diam}(U_j) < 2r_0$, we use $B(x_j, \operatorname{diam}(U_j)) \subset B(x_j, 2r_0)$ and note that $\mu(B(x_j, \operatorname{diam}(U_j)))$ is finite (as $\mu$ is a Borel measure); however, by restricting to $\delta < r_0$ we ensure $\operatorname{diam}(U_j) < r_0$ and the ball condition applies. So fix $\delta \in (0, r_0)$. Then for every $j$:
\begin{align*}
\mu(U_j \cap E) \leq c \cdot (\operatorname{diam}(U_j))^s.
\end{align*}
[/step]
[step:Sum over the cover and take the infimum]
Since $\{U_j\}$ covers $E$, the sets $\{U_j \cap E\}$ cover $E$, so
\begin{align*}
\mu(E) \leq \sum_{j=1}^\infty \mu(U_j \cap E) \leq c \sum_{j=1}^\infty (\operatorname{diam}(U_j))^s.
\end{align*}
Dividing by $c > 0$:
\begin{align*}
\frac{\mu(E)}{c} \leq \sum_{j=1}^\infty (\operatorname{diam}(U_j))^s.
\end{align*}
This holds for every countable cover $\{U_j\}$ of $E$ with $\operatorname{diam}(U_j) \leq \delta$. Taking the infimum over all such covers:
\begin{align*}
\frac{\mu(E)}{c} \leq \mathcal{H}^s_\delta(E).
\end{align*}
[/step]
[step:Send $\delta \to 0$ and conclude]
Since $\mathcal{H}^s_\delta(E)$ is non-decreasing as $\delta \to 0^+$ (smaller $\delta$ restricts the class of admissible covers), and $\mathcal{H}^s(E) = \lim_{\delta \to 0^+} \mathcal{H}^s_\delta(E)$, the bound $\mu(E)/c \leq \mathcal{H}^s_\delta(E)$ for all $\delta \in (0, r_0)$ gives
\begin{align*}
\mathcal{H}^s(E) = \lim_{\delta \to 0^+} \mathcal{H}^s_\delta(E) \geq \frac{\mu(E)}{c} > 0.
\end{align*}
The strict inequality $\mu(E)/c > 0$ holds because $\mu(E) > 0$ and $c > 0$ by hypothesis.
[/step]
