In a bichromatically colored plane, is it always possible to construct any regular polygon such that all vertices are the same color?

Question

Let there be a plane $P$ where all points $(x,y)$ are colored either $red$ or $blue$. Is it always possible to construct every regular $n$-gon where all the vertices are the same color?

Some observations and things (That might possibly be helpful?):

1. There is always a line such that its endpoints and midpoint are the same colors. (I feel like this one could be useful for showing continuity when done an infinite number of times? idk)
2. This is true for the triangle and square case. (Presumably also the rectangle case but I don't even know how to go about solving that. Also I feel like it should be easily provable with hexagons but I don't have the time right now - will update later)
3. It is obvious that if there are only a finite amount of points of some color, then it is possible (I think).

Note: if you can't, a counterexample would be appreciated, as well as any other exceptions and reasons as to why.

Double Note: if you can, I would also like to know if you can just make a line of any length and any shape always. (edit: I have since realized this is not true. Say we wanted to be able to make a straight line of length 1. Consider a $red$ plane. Now we can color a square grid $blue$ s.t the side length of each square is 0.5 units. Then, at each intersection of the squares, replace the vertex with a hollow $blue$ circle of appropriate radius, where the points inside the circle are $red$. We cannot make a straight line segment of any color over a certain length.)

Answer 1

An answer to your "double note"

Start with this:

Colour $(\mathbb{R}\setminus\mathbb{Q})\times(\mathbb{R}\setminus\mathbb{Q})$ and $\mathbb{Q}\times\mathbb{Q}$ red,

and the other points blue.

Then any line segment (a line segment being more than just a point) which is monochromatic is a segment of a rational line, i.e. defined by $ax+by+c=0$, where $a$, $b$ and $c$ are rational and either $a\neq0$ or $b\neq0$.

In order to end up with a colouring where none of these is monochromatic,

we are going to colour some of the points on some of them (which are now red) back to blue,

with the following procedure:

There are countably many rational lines. Enumerate them as $T_n$, that is to say that $n\mapsto T_n$ is a bijection from $\mathbb{N}$.

Let $L_n=T_n\cap(\mathbb{Q}\times\mathbb{Q})$. We just want to look at rational points for this procedure.

Choose a dense subset $S_1$ of $L_1$, such that $L_1-S_1$ is also dense in $L_1$. Colour $S_1$ blue.

Choose a dense subset $S_2$ of $L_2$, such that $S_2$ does not intersect $L_1$ anywhere and such that $L_2-S_2$ is also dense in $L_2$. Colour $S_2$ blue.

Choose a dense subset $S_3$ of $L_3$, such that $S_2$ does not intersect $L_1\cup L_2$ anywhere and such that $L_3-S_3$ is also dense in $L_3$. Colour $S_3$ blue.

Continue this recursively. Every rational line segment will have both colours.