$K_6$ contains at least two monochromatic $K_3$ graphs.

Since $R(3,3)=6$ there is a monochromatic triangle $\Delta$. Let's say it's blue. Look at the other three vertices. If there is no red edge between them then we've found a second blue triangle, so suppose we have found a red edge $xy$, $x,y\notin\Delta$. If there are two blue edges from $x$ to $\Delta$ then we've found a second blue triangle, so assume there are two red edges from $x$ to $\Delta$. Similarly assume there are two red edges from $y$ to $\Delta$. But that means that there is a $z\in\Delta$ such that $xz$ and $yz$ are both red, so we've found a red triangle.

There is another counting argument that one can come up with. A possible quantity of interest is a $monochromatic$ $angle$. An angle is monochromatic if both its arms are of the same colour.

Using pigeonhole principle, we can see that, at each vertex of $K_6$, there are at least three edges of the same colour (and hence at least 4 monochromatic angles). This leads to the conclusion that there are at least $24$ $(=6$x$4)$ monochromatic angles in $K_6$. Also, we can see that every monochromatic triangle has $3$ monochromatic angles and every other triangle has exactly $1$ monochromatic angle.

We know that there are a total of $20$ triangles in $K_6$. Let there be $x$ monochromatic triangles. Then there must be $(20−x)$ non-monochromatic triangles.

Putting all of this together, we get:

$3x+(20−x)$$≥24$ which gives $x≥2$.

I really like this argument and feel that the idea of monochromatic angles is very understudied. There are more things one can prove using this quantity and I encourage you to do so :)