Proof.
(1) Suppose that $K$ is not polar. By theorem 4.1.17, $K$ carries a nonzero positive measure of finite energy; dividing it by its mass, we obtain a probability measure $\mu $ on $K$ of finite energy $I(\mu )$ for the Newtonian kernel $k(x,y)=|x-y|^{2-2n}/((2n-2)s_{2n-1})$ of subsection 4.1.1 in $\mathbb R^{2n}$. Its potential $U^\mu $ is lower semicontinuous, and it is finite $\mu $-almost everywhere because $\int U^\mu \, \mathrm d\mu =I(\mu )<\infty $. The sets $E_ M=K\cap \{ U^\mu \le M\} $, $M\in \mathbb N$, are compact, because $U^\mu $ is lower semicontinuous, and they increase to a set of full $\mu $-measure; hence $\mu (E_ M)>0$ for $M$ large enough. Let $\nu $ be the restriction of $\mu $ to $E_ M$. For $a\in E_ M$ and $r>0$,
\[ \frac{r^{2-2n}\, \nu (B(a,r))}{(2n-2)s_{2n-1}}\leq \int _{B(a,r)}k(a,y)\, \mathrm d\nu (y)\leq U^\mu (a)\leq M, \]
so $\nu (B(a,r))\le C_ nMr^{2n-2}$ with $C_ n=(2n-2)s_{2n-1}$. Now cover $E_ M$ by balls $B(c_ l,r_ l)$ with $r_ l<1$ and $\sum _ lr_ l^{2n-3}<\eta $, and discard those not meeting $E_ M$. Each remaining ball lies in $B(a_ l,2r_ l)$ for some $a_ l\in E_ M$, so
\[ \nu (E_ M)\leq \sum _ l\nu (B(a_ l,2r_ l))\leq 2^{2n-2}C_ nM\sum _ lr_ l^{2n-2}\leq 2^{2n-2}C_ nM\eta , \]
where we used $r_ l<1$. Since $\eta >0$ is arbitrary, $\nu (E_ M)=0$, a contradiction. Hence $K$ is polar.
(2) Fix $x$. For $m\geq 1$ the set $K_ m=K\cap \{ |y-x|\geq 1/m\} $ is compact and $\mathcal H^{2n-3}$-null, and $K\setminus \{ x\} =\bigcup _ mK_ m$. The direction map $\Phi (y)=[y-x]$ is Lipschitz on $K_ m$. Indeed, $\Phi (y)=[(y-x)/|y-x|]$. The normalization $y\mapsto (y-x)/|y-x|$ has Lipschitz constant at most $2m$ on $\{ |y-x|\geq 1/m\} $: for nonzero vectors $a,b$,
\[ \Bigl |\frac a{|a|}-\frac b{|b|}\Bigr |\leq \frac{|a-b|}{|a|}+|b|\, \Bigl |\frac1{|a|}-\frac1{|b|}\Bigr |=\frac{|a-b|+\bigl ||b|-|a|\bigr |}{|a|}\leq \frac{2|a-b|}{|a|}, \]
and we apply this with $a=y-x$, $b=y'-x$ and $|a|\geq 1/m$. The Hopf map is Lipschitz with constant $\pi /2$ for the metric $d$. Indeed, let $e,e'$ be unit vectors and $\delta =\arccos \operatorname{Re}\langle e,e'\rangle \in [0,\pi ]$ their spherical distance. Since $d$ is the minimum of the spherical distance over the unit representatives (subsection 2.9.17), we have $d([e],[e'])\leq \delta $. Moreover $|e-e'|^2=2-2\cos \delta =4\sin ^2(\delta /2)$, and $\sin s\geq 2s/\pi $ for $0\leq s\leq \pi /2$, so $|e-e'|\geq 2\delta /\pi $. Hence
\[ d([e],[e'])\leq \delta \leq \tfrac \pi 2|e-e'|. \]
Since Lipschitz images of $\mathcal H^{2n-3}$-null sets are $\mathcal H^{2n-3}$-null, the compact set $\Phi (K_ m)$ is $\mathcal H^{2n-3}$-null in $(\mathbb P^{n-1}(\mathbb C),d)$. Cover it by balls of radii $r_ l<1$ with $\sum _ lr_ l^{2n-3}<\eta $. By lemma 2.9.19, and since $r_ l<1$, we have $\sigma (\Phi (K_ m))\leq \sum _ lr_ l^{2n-2}<\eta $, so $\sigma (\Phi (K_ m))=0$. Hence the set of directions $\Phi (K\setminus \{ x\} )=\bigcup _ m\Phi (K_ m)$ is a Borel set of $\sigma $-measure zero.
(3) An orthogonal projection onto a complex hyperplane $H\cong \mathbb R^{2n-2}$ is $1$-Lipschitz. Hence the projection of $K$ is $\mathcal H^{2n-3}$-null in $H$, and therefore Lebesgue null because $2n-3<2n-2$.
$\square$
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