Projection of $F$ onto any line nonparallel to coordinate axes has measure zero. Given a square $K=[a,b] \times [c,d]$, we define a union of four squares $K^*\subset K$ as follows: $I_i=[a_i, a_{i+1}]$, $J_i=[c_i, c_{i+1}]$, where $a_i=a+i\frac{b-a}{4}$, $c_i=c+i\frac{d-c}{4}$, for $i=0,1,2,3$ and we let $$K^*=(I_0\times J_1)\cup(I_1 \times J_0) \cup (I_2\times J_3)\cup(I_3 \times J_2).$$ We define a decreasing sequence of compact sets $F_0, F_1, \ldots$ inductively: $F_0=[0,1]^2$, $F_n$ is a finite union of squares $K$, each two having at most one point in common, and $F_{n+1}$ is obtained from $F_n$ by replacing each square $K$ by $K^*$. Let $F=\bigcap_{n=0}^\infty F_n$. 

Let $\mathcal H^1$ be the Hausdorff measure. Let $\ell$ be a line through $(0,0)$ different from the coordinate axes. Let $\pi:\mathbb R^2\to\ell$ be the orthogonal projection onto $\ell$. Show that $\pi(F)$ is of $\mathcal H^1$-measure null.
I have trouble with this exercise. I tried to calculate $\mathcal H^1(\pi(F_n))$ but I failed. Any hints?

EDIT: Theorem 3.32 in Falconer's "The geometry of fractal sets" states that the projection of $F$ onto almost every line has measure zero. This is weaker than the thesis of this exercise and the proof uses some non-trivial results. I am looking for an elementary argument.
 A: A partial answer, giving the result for some countable set of lines.
For any $n$ write $F_n = \bigcup_{i=1}^{4^n} K_i^n$ where $K_1^n, K_2^n, \ldots, K_{4^n}^n$ are disjoint squares with sides equal to $4^{-n}$. Consider the set $\mathcal L$ of lines $\ell$ with the property that for some $n, i, j$ with $i \ne j$ we have $\pi_\ell(K_i^n)=\pi_\ell(K_j^n)$. 
Fix $\ell \in \mathcal L$ and $n,i,j$ such that $i \neq j$ and $\pi_\ell(K_i^n)= \pi_\ell(K_j^n)$. Observe that $\pi_\ell(K_i^n\cap F) = \pi_\ell(K_j^n \cap F)$. Obviously, for any $k$ we have $\mathcal H^1(\pi_\ell(F \cap K_k^n))=4^{-n} \mathcal H^1(\pi_\ell(F))$. Thus $$\mathcal H^1(\pi_\ell(F)) = \mathcal H^1\left(\pi_\ell\left(\bigcup_{k=1}^{4^n} K_k^n\cap F\right)\right) = \mathcal H^1\left(\bigcup_{k=1}^{4^n} \pi_\ell(K_k^n\cap F)\right) \le \\ \le \mathcal H^1 (\pi_\ell(K_i^n\cap F)) + \sum_{k\notin\{i,j\}}\mathcal H^1 (\pi_\ell(K_k^n\cap F)) = (4^n-1)4^{-n}\mathcal H^1 (\pi_\ell(F)).$$ It follows that $H^1 (\pi_\ell(F))=0$.
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I think this may lead to a full proof. I suspect that the set $\mathcal L$ is dense (meaning that the set of angles between $\ell \in \mathcal L$ and the $x$-axis is dense in $[0,\pi]$). For any line $\mathcal k \notin \mathcal L$ we should be able to choose $\ell \in \mathcal L$ close enough and $n \in \mathbb N$ big enough to ensure that the measure of $\pi_{\mathcal k}(F_n)$ is very close to the measure of $\pi_{\ell}(F_n)$ which is close to zero. I did not have time to formalize this argument. 
