First we show that there exists either a monochromatic equilateral triangle of length $1$, or a monochromatic equilateral triangle of length $\sqrt{3}$. Indeed, if there is no monochromatic equilateral triangle of length $1$, then we can find two points $A$ and $B$ such that $AB = 1$ and they are in different colours (say $A$ is red and $B$ is blue). Construct an isosceles triangle $ABC$ with $AC = BC = 2$.

Without loss of generality, assume $C$ is blue. Let $M$ be the midpoint of $AC$. Then it has the same colour as $A$ or $C$, say $A$. Construct two equilateral triangles $ADM$ and $AEM$. Since there is no monochromatic equilateral triangle of length $1$, the colours of $D$ and $E$ are different from that of $A$, and hence both are blue. Thus $\triangle CDE$ is a monochromatic equilateral triangle of length $\sqrt{3}$.

Now suppose there is a monochromatic equilateral triangle $PQR$ of length $1$. We assume all its vertices are red in colour and construct three more equilateral triangles $PQW$, $QRU$ and $RPV$. If all of the points $U$, $V$, $W$ are blue, then $UVW$ is a monochromatic equilateral triangle of length $2$ and we are done. If not, at least one of them is red, say $U$, then $PQUR$ is a monochromatic rhombus of length $1$.