You only need to consider the case $\mathfrak{h}_{s}^\ast(A) \lt \infty$, but you need to be a bit careful in choosing the outer approximations since swapping $\inf$ and $\sup$ certainly isn't allowed without some thinking. If you knew that you can always take the same set $E$ in the $\inf$ (which I will show in \eqref{eqn:ast} below) then you'd be essentially done, but it is impossible to say if you got so far or not.
I'm going to ignore $\alpha_s$ since it only is a scalar factor that doesn't play a rôle in the argument and the formulas will already get cumbersome enough without it.
The following argument is essentially the one from Fremlin's measure theory, Volume 4II, 471D, page 102. $\DeclareMathOperator{\diam}{diam}$ $\newcommand{\hsd}[1]{\mathfrak{h}_{s,#1}^\ast}$
Set $\delta_n = 2^{-n}$ and choose $(A_{i}^n)_{i \in \mathbb{N}}$ of diameter $\leq 2^{-n}$ such that $A \subset \bigcup_{i=1}^\infty A_{i}^n$ and that
\begin{equation}\tag{$1$}\label{eqn:eq1}
\sum_{i=1}^\infty (\diam{A_{i}^n})^s \leq \hsd{2^{-n}}(A) +2^{-n},
\end{equation}
which is possible because of the definition of $\hsd{2^{-n}}$ as infimum.
Observe that there's no reason for the $A_{i}^n$ to be $\hsd{2^{-n}}$-measurable, let alone Borel, so we would like to “blow them up” slightly, so as to get open sets still approximating the $\hsd{2^{-n}}$-measure of $A$ well. The problem is that by doing so we will lose the diameter condition which appears in the definition of $\hsd{2^{-n}}$, but this isn't a serious problem: we can simply choose a larger $n$ and work with the sets we obtain from there. Here are the gory details:
Choose $0 \lt r_{i}^n \lt 2^{-n}$ so small that
$$
(\diam{(A_{i}^n)} + 2r_{i}^n)^s \leq (\diam{A_{i}^n})^s + 2^{-n-i}.
$$
Now put $U_{i}^n = \{x \in X\,:\,d(x,A_{i}^n) \lt r_{i}^n\}$ and note that $U_{i}^n \supset A_{i}^n$ is an open set whose diameter satisfies
\begin{equation}\tag{$2$}\label{eqn:eq2}
(\diam{U_{i}^n})^s \leq (\diam{A_{i}^n})^s + 2^{-n-i}.
\end{equation}
Let
$$
B = \bigcap_{n=1}^\infty \bigcup_{i=1}^\infty U_{i}^n
$$
and observe that $B$ is a $G_{\delta}$-set (countable intersection of open sets) containing $A$.
Given any $\delta \gt 0$, we can find $N$ such that $3 \cdot 2^{-N} \leq \delta$ so that for all $n \geq N$ we have $\diam{U_{i}^n} \leq \delta$. As $B \subset \bigcup_{i = 1}^\infty U_{i}^n$ we see from \eqref{eqn:eq1} and \eqref{eqn:eq2} that for $n \geq N$
$$
\hsd{\delta}(B) \leq \sum_{i=1}^\infty (\diam{U_{i}^n})^s
\leq
\sum_{i=1}^\infty [(\diam{A_{i}^n})^s + 2^{-n-i}] \leq \hsd{2^{-n}}{(A)} + 2 \cdot 2^{-n}
$$
Since $\hsd{2^{-n}}(A) \leq \mathfrak{h}_{s}^\ast(A)$ we get for $n \geq N$
$$
\hsd{\delta}(B) \leq
\mathfrak{h}_{s}^\ast(A) + 2^{-n}
$$
and letting $n \to \infty$ this gives
\begin{equation}\tag{$\ast$}\label{eqn:ast}
\hsd{\delta}(B) \leq \mathfrak{h}_{s}^\ast(A)
\end{equation}
for every $\delta \gt 0$. [Note: this is stronger than your condition involving $\inf$ since we can specify the set $E = B$ and thus avoid the infimum]
Taking the $\sup$ over all $\delta$ in \eqref{eqn:ast} this yields
$$
\mathfrak{h}_{s}^\ast(B) \leq \mathfrak{h}_{s}^\ast(A)
$$
and since $B \supset A$ and $B$ is $\mathfrak{h}_{s}$-measurable (being Borel) we can finally conclude
that
$$
\mathfrak{h}_{s}(B) = \mathfrak{h}_{s}^\ast(A),
$$
as desired.
What does $\rho (x,y)$ denote?
It denotes $d(x,y)$. That is, you have a typo somewhere: $d$ and $\rho$ are the same thing, the metric.
What is meant by the diameter of a set?
What you wrote: the supremum of pairwise distances between the points in your set. To develop intuition, draw a few shapes on the plane (which is an excellent example of a metric space) and determine their diameters. The diameter of a circle is just that, the diameter (hence the name). The diameter of a triangle is the length of its longest side. And so on.
I'm just trying to understand the intuition behind this definition.
The intuition is that if the object is $d$-dimensional, then $r^d$ roughly represents the volume of its piece of size $r$. Summing over all pieces, we should get something that is no less than the volume of the object. That is, the sums should not be arbitrarily close to zero. And if they are, then the value of $d$ we picked is too high, and the actual dimension of the object is lower than that. So we make $d$ smaller (i.e., take infimum over $d$).
The preceding paragraph is a lie, but this is what you get when you ask for intuition.
Best Answer
Notice that at scale $\delta$, we have that
$$\sum_{i = 0}^{\infty} \operatorname{diam}(U_i)^2 \le \delta \sum_{i = 0}^{\infty} \operatorname{diam}(U_i)$$
so that the sum of squares is much smaller than the sum without squares. It's never actually zero, but letting $\delta \to 0$ shows that any set with finite $\mathcal{H}^1$ measure zero has zero $\mathcal{H}^2$ measure. A minor refinement of the argument gives that finite $\mathcal{H}^s$ measure implies zero $\mathcal{H}^t$ measure whenever $s < t$.