For any point $P$ in the plane, denote by its colour by $c(P)$. 1. If all the points $Q$ on the circle of centre $P$ and radius $1$ have $c(Q) = c(P)$, we found both equilateral triangles of side $1$ and of side $\sqrt{3}$, monochromatic $c(P)$. 2. So there exist points $P$ and $Q$ with $PQ=1$ and $c(P)\neq c(Q)$. Consider a point $R$ with $PR=QR=2$; then either $c(R) \neq c(P)$ or $c(R) \neq c(Q)$; wlog say the latter. 3. So we have points $Q$ and $R$ with $QR=2$ and $c(Q)\neq c(R)$. Consider the midpoint $M$ of $QR$; then either $c(M) = c(Q)$ or $c(M) = c(R)$; wlog say the latter. 4. Finally, consider the points $S \neq T$ with $SM=SR=1=TM=TR$; if $c(S) = c(R)$, respectively $c(T) = c(R)$, we found the equilateral triangle $SMR$, respectively $TMR$, of side $1$ and monocromatic $c(R)$, while otherwise $c(S) \neq c(R)$ and $c(T) \neq c(R)$, so $c(S) = c(T) = c(Q)$ and we found the equilateral triangle $QST$ of side $\sqrt{3}$ and monocromatic $c(Q)$.