Elementary proof of corollary of Besicovitch projection theorem

Question

I am interested in the following fact.

---

Fact. Suppose

1. $\mathcal H^1(E) < \infty$.
2. $|\pi_\theta(E)| = 0$ for a.e. $\theta \in [0,\pi]$.
3. $\Psi : \mathbb{R}^2 \to \mathbb{R}^2$ is a $C^1$ map with nonvanishing Jacobian.

Then $|\pi_\theta(\Psi(E))| = 0$ for a.e. $\theta \in [0,\pi]$.

(Notation: $\mathcal H^1$ denotes Hausdorff measure. $\pi_\theta : \mathbb R^2 \to \mathbb R$ denotes the orthogonal projection in direction $\theta$. $|\cdot|$ denotes Lebesgue measure.)

---

The fact above is a simple consequence of the Besicovitch projection theorem and the fact that if $E$ is purely unrectifiable, then so is $\Psi(E)$. Since the Besicovtich projection theorem is quite difficult, I was curious if there is an "easier" proof.

Remark: Tuomas Orponen pointed out to me that assumption 1 cannot be removed because of the following result by Mattila:

Mattila, Pertti, Smooth maps, null-sets for integralgeometric measure and analytic capacity, Ann. Math. (2) 123, 303-309 (1986). ZBL0589.28006.