We first show that there exists a rectangle with monochromatic vertices, then subdivide it into $n^2$ rectangles by subdividing each side into $n$ congruent segments, and show that (at least) one of these smaller rectangles has (at least) three monochromatic vertices – here is where the assumption that $n$ be odd comes in.
To find a rectangle with monochromatic vertices, consider a line $a$ in the plane and a five-element monochromatic subset $A$ of $a$. Project $A$ orthogonally on another line $b$ which is parallel to $a$, and consider a three-element monochromatic subset $B$ of the image of $A$ under projection. If the colours of $A$ and $B$ agree, we are done. Otherwise, project $B$ on a third line $c$ which is parallel to both $a$ and $b$, and consider a two-element monochromatic subset $C$ of the image of $B$ under projection. Clearly, the two points of $C$ together with their orthogonal projections on either $A$ or $B$ are monochromatic.
Without loss of generality, we may (and will) assume that the vertices of the square $[0, n] \times [0, n]$ are monochromatic. Since $n$ is odd, and the points $0 \times 0$ and $n \times 0$ (respectively, $0 \times n$) are monochromatic, there exists $p$ (respectively, $q$) in $\{0, 1, \dots, n-1\}$ such that the points $p \times 0$ and $(p+1) \times 0$ (respectively, $0 \times q$ and $0 \times (q+1)$) be monochromatic. We now show that (at least) one of the unit squares
$$[i, i+1] \times [q, q+1], \quad i = 0, 1, \dots, p,$$
$$[p, p+1] \times [j, j+1], \quad j = 0, 1, \dots, q,$$
has (at least) three monochromatic vertices. Suppose that each of the first $p$ (respectively, $q$) horizontal (respectively, vertical) unit squares above has two vertices of each colour. Then the colours of the lattice points alternate in the same way along both horizontal segments $[0, p] \times q$ and $[0, p] \times (q+1)$, according to the parity of the abscissae, and in the same way along both vertical segments $p \times [0, q]$ and $(p+1) \times [0, q]$, according to the parity of the ordinates. Consequently, the vertical pair $(p \times q, p \times (q+1))$ and the horizontal pair $(p \times q, (p+1) \times q)$ are both monochromatic; that is, the unit square $[p, p+1] \times [q, q+1]$ has (at least) three monochromatic vertices.