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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,