Hardy–Littlewood maximal theorem

Step 1 (a cover by good balls). Fix $\lambda>0$. For every $x$ with $Mf(x)>\lambda$, by definition of the supremum there is some interval $B_x$ centered at $x$ with $\frac{1}{|B_x|}\int_{B_x}|f| > \lambda$. These balls $\{B_x\}$ cover the set $\{Mf>\lambda\}$.

Step 2 (Vitali $5r$-covering lemma). From any collection of balls of bounded radius, one can always extract a countable disjoint subcollection $\{B_i\}$ such that the $5$-times dilated balls $\{5B_i\}$ still cover the union of the whole original collection. Apply this to $\{B_x\}$ to get a disjoint subfamily $\{B_i\}$ with $\{Mf>\lambda\}\subseteq\bigcup_i 5B_i$.

Step 3 (sum the disjoint pieces). Each $B_i$ satisfies $\lambda|B_i| < \int_{B_i}|f|$ by construction. Since the $B_i$ are pairwise disjoint, summing over $i$ gives $\lambda\sum_i|B_i| < \sum_i\int_{B_i}|f| \le \int_{\mathbb{R}}|f| = \|f\|_1$, so $\sum_i|B_i| < \|f\|_1/\lambda$. Finally $|\{Mf>\lambda\}| \le \sum_i|5B_i| = 5\sum_i|B_i| < \dfrac{5}{\lambda}\|f\|_1$, which is exactly $|\{x : Mf(x) > \lambda\}| \le \frac{5}{\lambda}\|f\|_1$.