[proofplan]
The inequality $\underline{\dim}_M(E) \leq \overline{\dim}_M(E)$ is immediate from $\liminf \leq \limsup$. For $\dim_{\mathcal{H}}(E) \leq \underline{\dim}_M(E)$, we fix $s > \underline{\dim}_M(E)$ and show $\mathcal{H}^s(E) = 0$. We pick an intermediate exponent $\sigma \in (\underline{\dim}_M(E), s)$; the $\liminf$ condition ensures that for arbitrarily small radii $\rho$, the covering number satisfies $N(E, \rho) < \rho^{-\sigma}$. A cover by $N(E, \rho)$ balls of radius $\rho$ is a valid $2\rho$-cover for the Hausdorff computation, and the covering sum is bounded by $\alpha(s) \rho^{s - \sigma} \to 0$.
[/proofplan]
[step:Establish that $\underline{\dim}_M(E) \leq \overline{\dim}_M(E)$]
This is a general property of $\liminf$ and $\limsup$: for any function $g: (0, \infty) \to \mathbb{R}$,
\begin{align*}
\liminf_{r \to 0^+} g(r) \leq \limsup_{r \to 0^+} g(r).
\end{align*}
Applying this with $g(r) = \log N(E, r) / (-\log r)$ gives $\underline{\dim}_M(E) \leq \overline{\dim}_M(E)$.
[/step]
[step:Translate the condition $s > \underline{\dim}_M(E)$ into a covering bound]
Let $s > \underline{\dim}_M(E)$ and pick $\sigma$ with $\underline{\dim}_M(E) < \sigma < s$. By definition,
\begin{align*}
\underline{\dim}_M(E) = \liminf_{r \to 0^+} \frac{\log N(E, r)}{-\log r},
\end{align*}
where $N(E, r)$ is the minimum number of closed balls of radius $r$ needed to cover $E$. Since $\sigma$ exceeds the $\liminf$, for every $r \in (0, r_0)$ (for some $r_0 > 0$ depending on $\sigma$) there exists $\rho \in (0, r]$ with
\begin{align*}
\frac{\log N(E, \rho)}{-\log \rho} < \sigma, \qquad \text{equivalently } N(E, \rho) < \rho^{-\sigma}.
\end{align*}
(The $\liminf$ condition provides such $\rho$ for arbitrarily small $r$, though not necessarily for every $r$.)
For each $\delta > 0$, choose $r = \min(\delta/2, r_0)$ and then $\rho \in (0, r]$ with $N(E, \rho) < \rho^{-\sigma}$. Cover $E$ by $N(E, \rho)$ closed balls of radius $\rho$. Each ball has diameter $2\rho \leq 2r \leq \delta$, so this is a valid $\delta$-cover of $E$.
[/step]
[step:Bound $\mathcal{H}^s_\delta(E)$ using the covering by balls and send $\delta \to 0$]
Continuing from the previous step: fix $s > \underline{\dim}_M(E)$ and $\sigma \in (\underline{\dim}_M(E), s)$. For each $\delta > 0$ small enough, there exists $\rho \in (0, \delta/2]$ with $N(E, \rho) < \rho^{-\sigma}$. Cover $E$ by $N(E, \rho)$ closed balls $\{\overline{B}(x_j, \rho)\}_{j=1}^{N(E,\rho)}$. Each ball has $\operatorname{diam}(\overline{B}(x_j, \rho)) = 2\rho \leq \delta$, so this is a valid $\delta$-cover.
The Hausdorff covering sum is
\begin{align*}
\mathcal{H}^s_\delta(E) \leq \sum_{j=1}^{N(E,\rho)} \alpha(s) \left(\frac{2\rho}{2}\right)^s = N(E, \rho) \cdot \alpha(s)\, \rho^s < \rho^{-\sigma} \cdot \alpha(s)\, \rho^s = \alpha(s)\, \rho^{s - \sigma}.
\end{align*}
Since $s > \sigma$ and $\rho \leq \delta/2 \to 0$ as $\delta \to 0$, the bound $\alpha(s) \rho^{s-\sigma} \to 0$.
More precisely, for each $\delta > 0$, we have $\mathcal{H}^s_\delta(E) \leq \alpha(s) \rho^{s - \sigma}$ for some $\rho \leq \delta/2$. As $\delta \to 0$, we can choose $\rho \to 0$ (since $\rho \leq \delta/2$), so $\rho^{s - \sigma} \to 0$. Therefore
\begin{align*}
\mathcal{H}^s(E) = \lim_{\delta \to 0^+} \mathcal{H}^s_\delta(E) = 0.
\end{align*}
Since $\mathcal{H}^s(E) = 0$ for every $s > \underline{\dim}_M(E)$, the definition of Hausdorff dimension gives
\begin{align*}
\dim_{\mathcal{H}}(E) = \inf\{s \geq 0 : \mathcal{H}^s(E) = 0\} \leq \underline{\dim}_M(E).
\end{align*}
[/step]
