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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.

Label the edges so that barycentric coordinates on $T$ satisfySuppose for contradiction that $A\cap B\cap C=\varnothing$. The distance from a point to a closed set gives continuous nonnegative functions $d_A,d_B,d_C$. Their sum never vanishes, while at every point at least one of them vanishes because $A\cup B\cup C=T$. Henceis a continuous map from $T$ to its boundary in barycentric-coordinate space. Moreover, $F(I)\subseteq I$, $F(J)\subseteq J$, and $F(K)\subseteq K$. On each edge, the straight-line homotopy between $F$ and the identity remains in that edge, so $F{|}_{\partial T}$ has winding number one. On the other hand, a map from the whole triangle to its boundary makes its boundary restriction null-homotopic, and hence gives winding number zero. This contradiction proves the three-set covering lemma:

If a retraction $f:D\to \partial D$ fixed every boundary point, choose a homeomorphism from $T$ to $D$ taking its three edges to three consecutive closed arcs of $\partial D$ with empty triple intersection. The inverse images under $f$ of those arcs would be closed, would cover $D$, and would contain the corresponding boundary arcs. Transporting them to $T$ would contradict the result just proved. Therefore

For the final statement, use compactness of the three closed arcs inside the corresponding open sets. A finite open cover of a compact metric space admits a closed shrinking, and the shrinking can be chosen to preserve specified compact subsets already lying in the respective open sets. Thus there are closed setswhich cover $D$ and contain $\alpha ,\beta ,\gamma$, respectively. The disc version of the three-set covering lemma, obtained from the triangle by a homeomorphism taking its edges to the three arcs, gives a point in $A\cap B\cap C$. Since these sets lie in the original open sets,