Hello Martijn,

i guess the following argument should work:

For simplicity I shall assume that $E$ is bounded, i.e. there exists some ball $B_R(0)\subset \mathbb R^n$ strictly containing $E$. W.l.o.g. we may also assume that $H =$ {$x\in \mathbb R^n:x_n < r$} for some $r$. It holds the following (see the book of Giusti for details) $$P(E \cap H)=P(E,H)+\int_{\partial H}\phi_E^+ d\mathcal H^{n-1},$$ where $\phi_E^+$ denotes the inner trace of the characteristic function $\chi_E$ of $E$ on $\partial H$. Now by definition of the trace operator we have (inserting the vector field $X=-\eta e_n$, where $\eta$ is some smooth cutoff function that equals $1$ on $B_R(0)$) \begin{align} 0&=\int_{E\setminus \overline H}div X d\mathcal L^n=-\int_{\mathbb R^n \setminus \overline H} \langle e_n,\nu_E \rangle d\mu_E - \int_{\partial H}\phi_E^{-}\langle e_n,\nu_H\rangle d\mathcal H^{n-1}, \end{align} where $\nu_H$ denotes the outer unit normal to $H$, $\nu_E$ denotes the generalized outer unit normal of $E$ and $\phi_E^-$ is the outer trace of $E$ on $\partial H$. But since $\nu_H=e_n$ we get, using Cauchy-Schwarz' inequality: $$\int_{\partial H}\phi_E^- d\mathcal H^{n-1} \leq P(E,\mathbb R^n \setminus \overline H),$$ and so for a.e. value of $r$, $\phi_E^+=\phi_E^-$ $\mathcal H^{n-1}$-a.e. on $\partial H$, which yields $$P(E \cap H)\leq P(E,H)+P(E,\mathbb R^n \setminus \overline H)=P(E). $$ For an arbitrary value of $r$ just choose a sequence of values $r_k\to r$ for which this is satisfies. Then $E\cap H_{r_k} \to E \cap H$ in $L_{loc}^1(\mathbb R^n)$. Using the lower semi-continuity of the perimeter you get the claim for all values of $r$.