In practice, I prefer the (equivalent) following definition of upper Minkowski dimension:
$\overline{\dim}(E) = \limsup_{\delta \to 0} \frac{\log N(E,\delta)}{\log(1/\delta)}. $
$N(E, \delta)$ is the number of $\delta$-balls needed to cover the set $E$. (We need $E$ to be compact).
(Throughout, we write $X \approx Y$ if there is a constant $c\geq 1$ such that $\frac{1}{c}X \leq Y \leq c X$. We also write $X \sim Y$, if $\lim (X/Y) = 1$ with respect to some common paramater of $X$ and $Y$.)
EDIT: To see that these notions are equivalent, just notice that $\mu(E_{\delta}) \approx \delta^n N(E,\delta)$. Formally, for a compact set $E \subset \mathbb{R}^d$, we have
$\begin{align} \overline{\dim}_M(E) &= \inf\left\{\alpha \in [0,d] : \delta^{-n}\mu(E_{\delta}) \lesssim \delta^{-\alpha}\right\} \\ &= \inf \left\{\alpha \in [0,d] : \limsup_{\delta \to 0} N(E,\delta) \cdot \delta^{\alpha} = 0\right\} \\ &= \limsup_{\delta \to 0} \frac{\log N(E,\delta)}{\log (1/\delta)} \end{align}$
With this in mind, let's consider how many $\delta$-balls are necessary to cover $E = \{0,1, 1/2, \dots\}$. First, break up the set $E$ into two pieces:
$\left\{1, \frac{1}{2}, \dots , \frac{1}{n}\right\} \cup \left( \left\{\frac{1}{n+1}, \dots \right\} \cup \{ 0 \} \right) = A \cup B$
Well, we need $n$ $\delta$-balls to cover $A$ (There are $n$ distinct points), and $\approx n$ $\delta$-balls to cover $B$ (since $| \frac{1}{n+1} - 0| \cdot n \approx 1$). (We needn't be exact with the precise value of $N(E, \delta)$. We really only need an asymptotic value).
Therefore, $N(E, \delta) = n + n \approx n$. However, what is $n$ in terms of $\delta$? As long as $\delta \ll \frac{1}{n} - \frac{1}{n+1} \approx \frac{1}{n^2}$, we are not overlapping. Therefore, we can choose $\delta \approx \frac{1}{n^2}$ or $n \approx 1/\sqrt{\delta}$.
Finally:
$ \overline{\dim(E)} = \limsup_{\delta \to 0} \frac{\log(1/\sqrt{\delta})}{\log(1 /\delta)} = \frac{1}{2}. $
This argument can be made as formal as you like, though we really only need an asymptotic value for $N(E, \delta)$, and $\delta$ (since $\log cx \sim \log x$).
The trick with calculating the Minkowski dimension of a set is to let $\delta \to 0$ and $n \to \infty$ in a controlled way. Here, setting $\delta \approx \frac{1}{n^2}$ did the trick.
I would try adapting the argument to find the (upper) Minkwoski dimension of the set $E_{\alpha} = \{1, \frac{1}{2^{\alpha}}, \frac{1}{3^{\alpha}}, \dots \} \cup \{ 0 \}$ (here, \alpha > 0). I find these examples really fun to play with! (This also let's one construct a set with Minkowski dimension $\beta$ for any $0 < \beta < 1$).
Furthermore, if you are ever in need of a good textbook or reference on harmonic analysis and Hausdorff or Minkowski dimensions, I would strongly recommend Tom Wolff's "Lectures on harmonic analysis" and Pertti Mattila's "Geometry of Sets and Measures in Euclidean Spaces"(which is basically the gold standard for geometric measure theory!).