You do not need such heavy high-tech as Korn's inequality or even Lebesgue measure theory for an elementary geometry homework. Let's say $F(0)=0$.

The first claim is that $F$ is bounded in some neighborhood of the origin. Indeed, just choose finitely many points $x_j$ such that $x_i-x_j$ span the space. Then for any $x$ in some fixed ball we have $\langle F(x)-F(x_j), x-x_j\rangle$ bounded by some constant whence $\langle F(x), x-x_j\rangle$ are bounded by a constant, whence $\langle F(x), x_i-x_j\rangle$ are bounded too and that is the end of this story.

Now suppose $x_j\to 0$ with $|F(x_j)|\ge\delta>0$. Then WLOG $F(x_j)\to u\ne 0$. Take any $x\ne 0$ close to $0$. We have $$ |\langle F(x)-0,x-0\rangle|\le |x|\rho(|x|) $$ and $$ |\langle F(x)-F(x_j),x-x_j\rangle|\le |x-x_j|\rho(|x-x_j|) $$ which, after passing to the limit, becomes $$ |\langle F(x)-u,x\rangle|\le |x|\rho(|x|) $$ Subtracting, we get $$ |\langle u,x\rangle|\le|x|\rho(|x|) $$ but this is absurd if $x$ is collinear with $u$ and $\rho(|x|)<|u|$.

You do not need such heavy high-tech as Korn's inequality or even Lebesgue measure theory for an elementary geometry homework. Let's say $F(0)=0$.

The first claim is that $F$ is bounded in some neighborhood of the origin. Indeed, just choose finitely many points $x_j$ such that $x_i-x_j$ span the space. Then for any $x$ in some fixed ball we have $\langle F(x)-F(x_j), x-x_j\rangle$ bounded by some constant whence $\langle F(x), x-x_j\rangle$ are bounded by a constant, whence $\langle F(x), x_i-x_j\rangle$ are bounded too and that is the end of this story.

Now suppose $x_j\to 0$ with $|F(x_j)|\ge\delta>0$. Then WLOG $F(x_j)\to u\ne 0$. Take any $x\ne 0$ close to $0$. We have $$ |\langle F(x)-0,x-0\rangle|\le |x|\rho(|x|) $$ and $$ |\langle F(x)-F(x_j),x-x_j\rangle|\le |x-x_j|\rho(|x-x_j|) $$ which, after passing to the limit, becomes $$ |\langle F(x)-u,x\rangle|\le |x|\rho(|x|) $$ Subtracting, we get $$ |\langle u,x\rangle|\le 2|x|\rho(|x|) $$ but this is absurd if $x$ is collinear with $u$ and $2\rho(|x|)<|u|$.