 Solution

The triangular-grid form is Sperner's lemma: subdivide a triangle into small triangles meeting edge to edge, label its vertices $1,2,3$, and forbid label $i$ on the side opposite corner $i$. The corner labels are thereby forced. An odd number of small triangles, in particular at least one, carry all three labels.

Here is the parity proof. Count incidences of a small triangle with an edge whose endpoint labels are $1,2$. On the outer boundary these edges occur only on the side joining corners $1,2$; traversing that side begins with label one and ends with label two, so the number of changes of label is odd. Interior edges contribute twice to the incidence count. A small triangle with all three labels contributes once; a triangle having only labels one and two contributes twice if it has both, and all other triangles contribute zero. Reduction modulo two therefore proves the assertion.

To obtain the continuous Brouwer fixed-point theorem on the triangle, use barycentric coordinates and a continuous map $F$ from the triangle to itself. If a grid vertex is not already fixed, label it by an index $i$ with $x_i>F_i(x)$, which exists since both coordinate sums are one. A point on the side $x_i=0$ cannot receive label $i$, so the discrete lemma applies. On successively finer grids take a fully labelled small triangle. Compactness gives a subsequence converging to one point $x$; all three vertices converge to it because the mesh tends to zero. Passing to the limit in the three label inequalities gives $x_i\ge F_i(x)$ for every $i$. Equal coordinate sums force equality throughout, hence $F(x)=x$.

A closed disk is homeomorphic to a triangle: radial coordinates from an interior point, scaled by the distance to the boundary in each direction, give such a homeomorphism. Conjugation therefore gives the same fixed-point assertion on the disk. If $r$ were a continuous retraction from the unit disk to its boundary, define $F(x)=-r(x)$. A fixed point would have $|x|=1$, hence $r(x)=x$ and $x=-x$, impossible on the unit circle. Thus no continuous disk-to-boundary retraction exists. The continuous conclusion has been deduced from the grid parity argument, not used to prove it.