Let $E \subseteq \mathbb{R}^d$. Let $\epsilon > 0$. Show that there exist some countable covering $\{Q_k\}$ such that
$$|E|_e \leq \sum_k \text{vol}(Q_k) \leq |E|_e + \epsilon$$
I need a hint on how to proceed with such a proof, not necessarily a full solution. I think there is some property of infimums that lets me "squeeze" in numbers, but I don't know what that property is.
If $|E|_e = \infty$, pick a trivial covering that diverges. Otherwise, since
$$ |E|_e = \inf \left \{ \sum_{n \geq 1}\operatorname{vol}(Q_k) : \{Q_k\}_{k \geq 1} \text{ covers $E$ }\right\} $$
for any $\varepsilon > 0$, there exists some covering $\{Q_k\}_{k \geq 1}$ so that
$$ 0 \leq \sum_{n \geq 1}\operatorname{vol}(Q_k) - |E|_e < \varepsilon $$
which is exactly what you have to prove.
The property we are using is the following, assuming that you've defined the infimum as the 'larger' lower bound of a set:
Proof. Let $\varepsilon > 0$. Since $\iota < \iota + \varepsilon$, the latter cannot be a lower bound for $A$ because $\iota$ is an infimum. Thus, there exists $a \in A$ so that $a < \iota + \varepsilon$. Therefore,
$$ 0 \leq a - \iota < \varepsilon $$
as desired, with the first inequality given by the fact that $\iota$ is by construction less or equal than $a$. $\square$