Like usually in such situations, there is some doubt on the colouring of the separating lines. One idea would be for a unit square to be considered to be made of its interior, together with its left and lower sides, but less its north-west and south-east corners. This establishes a partition of the plane, given by
$$\bigcup_{(a,b) \in \mathbb{Z}^2} \{(x,y); a \le x < a+1,\ b \le y < b+1\}.$$

Another idea would be not to colour at all the sides of the unit squares – this has been chosen for the problem at hand. Moreover, this introduces the slight extra difficulty in avoiding the uncoloured points of the plane.

The key to the proof is the well-known configuration that warrants the existence of an equilateral monochromatic triangle in the bichromatic plane. Choose a value $\ell$ for the side of an equilateral triangle of integer area $\alpha$, hence $\ell = \sqrt{\frac{4\alpha}{\sqrt{3}}}$. Let $\gamma$ be the circle of radius $\ell$ and center some point $A$ coloured $c_1$, and $\Gamma$ the circle of same center and radius $\ell\sqrt{3}$. The meeting points of circles $\gamma$ and $\Gamma$ with the sides of the unit squares in the plane are obviously finitely many, and these points must be avoided since being uncoloured.

If all points of $\gamma$ (except at most a finite number of them) are coloured $c_2$, then there will exist an equilateral triangle whose vertices bear this colour, and of area $3\alpha$. Otherwise, let $B \in \gamma$ be coloured $c_1$, and consider the regular hexagon $BCDEFG$ inscribed in $\gamma$. Its vertices must bear colours $C \to c_2$, $G \to c_2$, $E \to c_1$, $D \to c_2$, $F \to c_2$, otherwise an equilateral monochromatic triangle of area $\alpha$ or $3\alpha$ is made. Let now $H = BC \cap DE$; clearly $H \in \Gamma$. If $H$ is coloured $c_2$, then $\triangle CDH$ is monochromatic $c_2$ and has area $\alpha$; if $H$ is coloured $c_1$, then $\triangle BEH$ is monochromatic $c_1$ and has area $4\alpha$.