Define the diameter of a subset $Y \subseteq \mathbb R^n$ of the metric space $(\mathbb R^n, d)$ with the standard metric $d$ to be
$$\operatorname{diam}(Y) := \sup_{\mathbf x,\mathbf y \in Y} d(\mathbf x, \mathbf y).$$
Now define the ball $B_r(\mathbf y) := \{ \mathbf x \in \mathbb R^n : d(\mathbf x, \mathbf y) \leq r\}$ and consider the ball
$A := B_r(\mathbf 0)$ and a cover $\{A_k\}_{k\geq 1} \supset A$, which don't have to be optimal (a cover here only need to satisfy $A\subset \bigcup_{k\geq 1} A_k$).
The actual claim we proved in class was that the (outer) Hausdorff $n$-dimensional measure (here: $n\in\mathbb N$) is defined on the same $\sigma$-algebra as the $n$-dimensional Lebesgue measure, and that they agreed up to a constant $C(n)$ depending on $n$.
Doing so, we used the inequality
$$\displaystyle\sum_{k\geq 1} \operatorname{diam}(A_k)^n \geq (2r)^n$$
valid for any cover $\{A_k\}_{k\geq 1}$ to $A$ satisfying
$$\operatorname{diam}(A_k) < 2r \; \; \forall k$$
How can one prove this inequality?