Improving the constant in Hardy-Littlewood maximal inequality from $3^d$ to $2^d$ This is Excercise 42 in Terry's notes on Differentiation theorem. I find it interesting but stumbled to get what he meant in his hint.
The Hardy-Littlewood maximal inequality for an absolutely integrable function $f$ reads:
$$ m(\{Mf(x)\geq \lambda\}) \leq \frac{C_d}{\lambda}\int_\mathbb{R^{d}}|f(t)|dt, $$
where $m$ is the Lebesgue measure, $\lambda>0$ and $Mf(x)$ is the Hardy-Littlewood maximal function. Using a Vitali type of covering lemma one can show the constant can be $3^d$. The exercise asks to improve this to $2^d$, by noticing that $2$-scaled balls cover the centers. Terry hinted that one need to do some $\epsilon$ adjustment. However, I failed to see why and how can small adjustments to $2$-scaled balls (or perhaps something else) can still lead to a sufficient cover, given that choice of covering balls seems to be fixed.
What am I missing here?

Thanks to discussion with Del and Po-lam, I realised that Tao did't mean to use the original finite covering balls (those to be dilated by a magnitude of $3$) in the Vitali covering lemma.
Once realising this, one tries to realize the following (credit to Yung Po-lam and Ding Cong ), see also the comments below for some further hint:
Suppose $K$ is a compact set, and for every $x \in K$, we are given an
open ball $B(x,r_x)$ that is centered at $x$ and of radius $r_x$.
Assume that $$R:= \sup_{x \in K} r_x < \infty.$$ Let $\mathcal{B}$ be
this collection of balls, i.e. $$\mathcal{B} = \{B(x,r_x) \colon x \in
K\}.$$ Then given any $\varepsilon > 0$, there exists a finite
subcollection $\mathcal{C}$ of balls from $\mathcal{B}$, so that the
balls in $\mathcal{C}$ are pairwise disjoint, and so that the
(concentric) dilates of balls in $\mathcal{C}$ by $(2+\varepsilon)$
times would cover $K$.
