Steinhaus follows in two lines from the density theorem

By [L1], choose a density-one point $x\in E$. Then there is $r>0$ such that [L1, L3, given, choose] $\lambda (E\cap B(x,r))>\frac{3}{4}\lambda (B(x,r))\mathrm{.}$ Because [L3] makes the ball measure continuous in the radius, choose $0<\delta <r$ with $\lambda (B(x,r+\delta )\setminus B(x,r-\delta ))<\frac{1}{4}\lambda (B(x,r))\mathrm{.}$ If $\mathrm{\parallel }h{\mathrm{\parallel }}_2<\delta$, then $B(x,r-\delta )\subseteq B(x+h,r)\subseteq B(x,r+\delta ),$ so $B(x+h,r)\mathrm{△}B(x,r)\subseteq B(x,r+\delta )\setminus B(x,r-\delta )\mathrm{.}$ Therefore $\lambda (B(x+h,r)\mathrm{△}B(x,r))<\frac{1}{4}\lambda (B(x,r))\mathrm{.}$

Fix $\mathrm{\parallel }h{\mathrm{\parallel }}_2<\delta$. Translation invariance [L2] gives [L2, step 1.1, algebra] $\lambda ((E-h)\cap B(x,r))=\lambda (E\cap B(x+h,r))\ge \lambda (E\cap B(x,r))-\lambda (B(x+h,r)\mathrm{△}B(x,r))>\frac{1}{2}\lambda (B(x,r))\mathrm{.}$ Together with step 1.1, both $E\cap B(x,r)$ and $(E-h)\cap B(x,r)$ have measure greater than half of $\lambda (B(x,r))$, so they intersect. Choose $z\in E\cap (E-h)$. Then $z\in E$ and $z+h\in E$, so $h=(z+h)-z\in E-E$.

Every $h$ with $\mathrm{\parallel }h{\mathrm{\parallel }}_2<\delta$ lies in $E-E$, so $E-E$ contains the open [step 2.1] ball $B(0,\delta )$.