Lebesgue density theorem in the line
Here is a nice proof due to Sierpiński, Démonstration élémentaire du théorème sur la densité des ensembles, Fundamenta Mathematicae, 4 (1), (1923), 167-171. I learned it from Appendix D in van Rooij, and Schikhof, A second course on real functions.
The argument is indeed elementary, and applies to all sets, even non-measurable ones:
Theorem. Given $E\subseteq\mathbb R$, almost every point of $E$ is an exterior density point of $E$, that is, for almost every $a\in E$, we have $$ \lim_{r\to0^+}\frac{m^*(E\cap(a-r,a+r))}{2r}=1, $$ where $m^*$ denotes Lebesgue outer measure.
The usual version follows, since if $E$ is measurable, then so is $E\cap(a-r,a+r)$, and so its outer measure is just its measure.
To prove the result, note that we may assume that $E$ is bounded. It is enough to show that if $t\in(0,1)$, then $$ A=\left\{a\in E\mid \liminf_{r\to0^+}\frac{m^*(E\cap(a-r,a+r))}{2r}<1-t\right\} $$ is null. (Recall that if $g$ is defined on $(a,a+\eta)$, then $$\liminf_{y\to a^+}g(y)=\sup_{0<\epsilon<\eta}\inf\{g(y):a<y<a+\epsilon\}.)$$
Fix $\epsilon>0$. We show that $m^*(A)<\epsilon(1+3/t)$. Since $t$ is fixed, this gives the result. To prove this, start by fixing $U$ open covering $A$ and of measure $m(U)<m^*(A)+\epsilon$ (which exists, by definition of outer measure, and the fact that $A$ is bounded). Note first that if $X\subseteq U$ is measurable, then $$ \begin{array}{cl} m^*(A)&\le m^*(A\cap X)+ m^*(A\setminus X)\le m^*(A\cap X)+m(U\setminus X)\\ &=m^*(A\cap X)+m(U)-m(X)\le m^*(A\cap X)+m^*(A)+\epsilon-m(X),\end{array} $$ so
$$m^*(A\cap X)\ge m(X)-\epsilon. $$
For each $a\in A$, pick an open interval $I$ with rational endpoints such that $a\in I\subseteq U$ and $m^*(E\cap I)<(1-t)m(I)$. This gives us a covering of $A$ by countably many open intervals $(I_n)_{n\in\mathbb N}$, all contained in $U$, and such that
$$ m^*(E\cap I_n)<(1-t)m(I_n) $$
for all $n$. Write each $I_n$ as $(x_n-\delta_n,x_n+\delta_n)$, and let $J_n=(x_n-3\delta_n,x_n+3\delta_n)$. Now, since $m^*(A)\le m(\bigcup_n I_n)$, we can find $N$ large enough that $$m^*(A)-\epsilon<m(\bigcup_{n\le N} I_n).$$ If needed, rearrange the intervals $I_1,I_2,\dots,I_N$ so $\delta_1\ge\delta_2\ge\dots\ge\delta_N$.
By a straightforward recursion, define a subset $L$ of $\{1,\dots,N\}$ with the property that, for each $i$ with $1\le i\le N$, the set $\{I_n\mid n\in L\cap[1,i]\}$ is parwise disjoint and maximal with this property. That $\{I_n\mid n\in L\cap[1,i]\}$ is maximal disjoint for each $i\le N$ implies that, for each $i\le N$, there is an $n\in L\cap[1,i]$ such that $I_n\cap I_i\ne\emptyset$. Since $\delta_n\ge\delta_i$, it follows that $I_i\subseteq J_n$. Therefore, $$ \bigcup_{i\le N}I_i\subseteq \bigcup_{n\in L}J_n. $$
Now set $X=\bigcup_{n\in L}I_n$ and note that, since the $I_n$ for $n\in L$ are disjoint, then $m(X)=\sum_{n\in L}m(I_n)$. This gives us that
$$ m^*(A)-\epsilon<m(\bigcup_{n\le N} I_n)\le m(\bigcup_{n\in L}J_n)\le\sum_{n\in L}m(J_n)=3\sum_{n\in L}m(I_n)=3m(X). $$
Since $$ m^*(E\cap I_n)<(1-t)m(I_n), $$ then
$$ m^*(A\cap X)\le\sum_{n\in L}m^*(A\cap I_n)\le \sum_{n\in L}m^*(E\cap I_n)\le\sum_{n\in L}(1-t)m(I_n)=(1-t)m(X).$$
But $$m^*(A\cap X)\ge m(X)-\epsilon, $$ so $m(X)\le\epsilon/t$ and, since $$ m^*(A)-\epsilon<3m(X),$$ then $$ m^*(A)<\epsilon+3\epsilon/t,$$ as we wanted.
(Note we are indeed using a covering result, namely that $\bigcup_{n\in L}J_n$ covers $\bigcup_{i\le N}I_i$. This is a particular instance of Vitali's covering lemma. A nice presentation of this and related covering results can be found in Krantz-Parks, The geometry of domains in space.)
This is nothing more than Lebesgue differentiation theorem applied to the function $f=\chi_A$. Follow this link to read the proof.
What about this excellent article in The American Mathematical Monthly?
They essentially only use properties of the outer measure.