 Solution

Sperner's lemma says that a triangulated triangle whose vertices are labelled $1,2,3$, with each outer vertex having its own label and each boundary edge using only its endpoint labels, contains a small triangle with all three labels. In fact the number of such triangles is odd. Count incidences of small triangles with edges labelled $1,2$. Interior edges contribute twice. On the outer $1,2$ edge, the number of changes between these labels is odd; the other two outer edges contribute zero. A small triangle contributes one incidence if it has all three labels, two if it uses both $1,2$ but not $3$, and zero otherwise. The total parity proves the assertion.

The resulting no-retraction theorem is: there is no continuous retraction of a closed disc onto its boundary. Work first with a closed triangle $P$. If $r:P\to \partial P$ fixes the boundary, colour each vertex of increasingly fine triangulations by an index of a largest barycentric coordinate of $r(v)$. The largest coordinate is at least $1/3$; on the boundary the colouring satisfies Sperner's lemma. Choose a tricoloured small triangle in each triangulation. By compactness and shrinking mesh, a subsequence of their three vertices converges to one point $x$. Continuity forces all three barycentric coordinates of $r(x)$ to be at least $1/3$, so $r(x)$ is the triangle's centre, contradicting $r(x)\in \partial P$. A homeomorphism of a disc with $P$ proves the disc assertion.

Brouwer fixed-point theorem states that every continuous self-map $F$ of a closed disc has a fixed point. If it had none, the ray from $F(x)$ through $x$ would meet the boundary beyond $x$, continuously in $x$. That intersection defines a retraction, since a boundary point is already the outgoing intersection. This contradicts the no-retraction theorem.

Finally fix $y\in {\mathbb{R}}^2$ and define $F_y(x)=y+x-g(x)$. The displacement bound implies $\Vert F_y(x)-y\Vert \le K$, so $F_y$ maps the closed radius-$K$ disc centred at $y$ into itself. Brouwer fixed-point theorem gives $F_y(x)=x$, which is exactly $g(x)=y$. Since $y$ was arbitrary, $g$ is surjective.