Def : In $\mathbb{R}^2$, $1$-set is a set that has 1-dimensional Hausdorff dimension and positive measure.
Def : Lower density and upper density of $F$ are $$\lim\ \inf_r\ H^1 (F\cap B(x,r))/2r ,\ \lim\ \sup_r\ H^1(F\cap B(x,r))/2r $$ where $H^1$ is a Hausdorff measure.
Def : $F$ is curve-free if $H^1 (F\cap c)=0$ where $c: [0,1]\rightarrow \mathbb{R}^2$ is an injective continuous map of finite length.
EXE : Let $F$ be a curve-free $1$-set. Then it has a lower density at $H^1$-almost all $x\in F$ which is $\leq 3/4$.
Proof : In general, upper density of $1$-set at almost all points is between $\frac{1}{2}$ and $1$.
How can we prove this ?
This is Theorem 3.23 in The Geometry of Fractal Sets by Falconer (the result is originally due to Besicovitch). Based on the length of the proof, it doesn't look like a good exercise to me. The proof relies on the following Lemma, in which $R(a,b) = B(a, r)\cap B(b, r)$ is the intersection of disks of radius $r=|a-b|$ centered at $a$ and $b$.
Lemma 3.22 Let $E$ be a $1$-set in $\mathbb{R}^2$ and suppose that $\alpha>0$. Let $E_0$ be a compact subset of $E$ with $H^1(E_0)>0$ such that $$H^1(E\cap R(a,b))\ge \alpha |a-b|\quad \forall \ a,b\in E_0$$ Then $E$ is not curve-free.
The proof of this lemma, including preliminary results, is about 3 pages (pages 40-43 in the book).
Assuming the lemma, the proof goes as follows. By inclusion-exclusion, $$ H^1(E\cap R(a,b)) + H^1(E\cap (B(a,r)\cup B(b, r))) = H^1(E\cap B(a,r)) + H^1(E\cap B(b,r)) $$ where $r=|a-b|$. If the lower density is greater than $3/4$ on a set of positive 1-measure, then by picking a small $\alpha>0$ and restricting $a,b$ to a suitable subset of $E$ we can get $H^1(E\cap B(a,r)) > (\frac34+\alpha)2 r$ and $H^1(E\cap B(b,r)) > (\frac34+\alpha)2 r$. On the other hand, the upper density is $\le 1$ a.e., and this is true even for the more general version of upper density, defined by $$ D_c^*(E, x) = \lim_{r\to 0} \sup_{x\in U, \ \operatorname{diam}U\le r} \frac{H^1(E\cap U)}{\operatorname{diam} U} $$ (It's called "upper convex density" in the book, but $U$ doesn't have to be convex here.) Applying this with $U=B(a,r)\cup B(b, r)$ yields $$ H^1(E\cap (B(a,r)\cup B(b, r))) \le (1+\alpha) \operatorname{diam}U = (1+\alpha) 3r $$ Putting it all together, $$ H^1(E\cap R(a,b)) \ge 4 \left(\frac34+\alpha\right) r - 3 (1+\alpha) r = \alpha r $$ allowing us to apply Lemma 3.22.
Aside: "purely unrectifiable" is a term often used for such sets.