For $d\ge 0$ and $S$ in a metric space $X$, we define the $d$-dimensional Hausdorff outer measure of $S$ as
$$ \mathcal{H}^d(S):=\lim_{r \to 0} \inf\Bigl\{\sum_i r_i^d:\text{ there is a cover of } S\text{ by balls with radii } 0 < r_i < r\Bigr\}. $$
By definition the Hausdorff dimension of $S$ is $\dim_{\operatorname{H}}(S):=\inf\{d\ge 0: \mathcal{H}^d(S)=0\}.$
I wonder why the following statement in Wikipedia is true:
This is the same as the supremum of the set of $d ∈ [0, ∞)$ such that the $d$-dimensional Hausdorff measure of $S$ is infinite (except that when this latter set of numbers d is empty the Hausdorff dimension is zero).
Namely, $\inf\{d\ge 0: \mathcal{H}^d(S)=0\}=\sup\{d\ge 0: \mathcal{H}^d(S)=\infty \}$. Clearly we have $\ge$, but why do we have $\le$?
Let $\alpha = \dim_H(S)$ (defined via the inf definition). As you say, it's clear that $\mathcal{H}^d(S) = 0$ for any $d > \alpha$ (since $\mathcal{H}^d(S)$ is a decreasing non-negative function of $d$).
So it suffices to show that $\mathcal{H}^d(S) = \infty$ for any $d < \alpha$. Let $$ \mathcal{H}^d_r(S) = \inf \left \{ \sum r_i^d : \text{ there is a cover of $S$ by balls $B_i$ with radii $r_i < r$ } \right\}. $$ Now let $r>0$ be small and consider any cover of $S$ by balls of radii $r_i < r$. We have $$ \sum r_i^d = \sum r_i^\alpha r_i^{d-\alpha} \geq r^{d-\alpha} \sum r_i^\alpha \geq r^{d-\alpha} \cdot \mathcal{H}^\alpha_r(S). $$ The first inequality is because $r<1$ and $d-\alpha < 0$. So $$\mathcal{H}^d_r(S) \geq r^{d-\alpha} \mathcal{H}^\alpha_r(S).$$ If $\mathcal{H}^\alpha(S) > 0$, then the right hand side tends to $\infty$ as $r \to 0$ because $d-\alpha < 0$, which implies that $\mathcal{H}^d(S) = \infty$.
Now consider the case where $\mathcal{H}^\alpha(S) = 0$. Since $\alpha$ is defined as the inf of the set where the Hausdorff content equals zero, necessarily we can find $d < \beta < \alpha$ such that $\mathcal{H}^\beta(S) > 0$ (again using monotonicity). Then just apply the above argument with $\beta$ in place of $\alpha$.