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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.

Sperner's lemma says that if a triangle is triangulated, its vertices are labelled $1,2,3$, and label $i$ is forbidden on the edge opposite vertex $i$, then an odd number of small triangles have all three labels.

To prove it, count incidences with edges whose endpoint labels are $1$ and $2$. Along the outer $12$-edge, the labels begin at $1$ and end at $2$, so the number of $12$ transitions is odd; no other boundary edge contributes. An interior edge contributes twice. A small triangle contributes an odd number precisely when its three labels are $1,2,3$: a triangle using only $1,2$ contributes two, and every other non-tricoloured triangle contributes zero. The number of tricoloured triangles is therefore odd.

Now suppose the closed sets $A_i$ in the question existed. Let $d_i(x)$ be the distance from a point to a closed set $A_i$. Empty triple intersection makes $d_1+d_2+d_3>0$, sois a continuous function from the large triangle to itself, where $v_i$ is the vertex opposite ${\alpha }_i$. Since the $A_i$ cover the triangle, at least one $d_i(x)$ vanishes, so $r(x)$ always lies on the boundary. On ${\alpha }_i\subseteq A_i$, its $i$th barycentric coordinate vanishes, hence $r({\alpha }_i)\subseteq {\alpha }_i$. Thus $r$ is face-preserving on the boundary; its boundary restriction is homotopic there to the identity by the straight-line homotopy. This would give a map from the triangle into its boundary whose boundary degree is one, contradicting the no-retraction theorem, the standard topological consequence of Sperner's lemma. Therefore the three closed sets must have a common point.

The equivalent no-retraction theorem says there is no continuous retraction of the closed disk onto its boundary circle. A fixed-point-free map gives a retraction by projecting from $f(x)$ through $x$ to the boundary; conversely a retraction composed with the antipodal boundary map gives a fixed-point-free self-map.

The planar Brouwer fixed-point theorem says every continuous $F:D^2\to D^2$ has a fixed point. The equivalent no-retraction theorem says no continuous $r:D^2\to S^1$ restricts to the identity on $S^1$. A fixed-point-free map gives a retraction by extending the ray from $F(x)$ through $x$ to the boundary; a retraction followed by the antipodal map gives a fixed-point-free self-map.

Suppose the stated sphere map is not onto and omits $p$. Compactness gives $\epsilon >0$ such that its image lies in $S^2\setminus B_{\epsilon }(p)$. That set is homeomorphic to a closed disc. Restricting $f$ to it gives a continuous self-map of a disc, so Brouwer supplies a fixed point, contrary to the hypothesis. Hence $f$ is surjective.