Let $X$ be a finite set of points in Euclidean plane. It is given that those points are not colinear. i.e. there is no line such that includes all points of $X$. Show that for any function $f \colon X \to \{0,1\}$(coloring the points with 2 color), there is a line $L$ on the plane such that satisfies
(a) $L$ contains at least 2 points in $X$. Namely, $|L \cap X| \geq 2$.
(b) $\forall a,b \in L \cap X$, $f(a) = f(b)$ (all points of $X$ on $L$ have same color).
I reduced this geometric problem in terms of equivalence relation. Let $Y$ be a set of 2-subsets of $X$, and define $Z = \{\{a,b\} \in Y | f(a) = f(b) \}$. Give equivalence relation $\sim$ on $Y$ to be $\{a,b\} \sim \{c,d\}$ iff $a,b,c,d$ are colinear. If $\mathcal{P}$ is a partition induced by $\sim$, then the problem is asking for the existence of $P \in \mathcal{P}$ such that $P \subseteq Z$.
When presenting this problem, my instructor said that combinatorial proof is very complicated, but there is a simple topological proof. Indeed I cannot figure out what to do after the reduction. What are the topological properties that I can exploit to attack this problem? How can I lead to a proof with those properties?
Thanks in advance for every help, hint, or solution.
[Inductive Proof Has a Problem in the Step Case]
Take $f:X\subseteq\mathbb{R}^2\to \{0,1\}$. As discussed, the cases where $f$ is identically $1$ or $0$ hold trivially. So proceed with the assumption that $f$ is surjective.
$\bullet\text{ }\underline{Base\text{ }Case\text{ }(n=3):}$
If $|X|=3$, $f$ surjective implies: $$\exists p,q,r\in X:\text{ }f(p) = f(q)\neq f(r).$$ Define: $$L_{p,q} := \{p+(q-p)t\text{ }|\text{ }t\in \mathbb{R}\}\subseteq \mathbb{R}^2.$$ Then $\exists L:= L_{p,q}$ such that $|L\cap X| = |\{p,q\}|=2$ and $f(p)=f(q)$. So $(a)$ and $(b)$ hold for $L$.
$\bullet\text{ }\underline{Hypothesis\text{ }Case\text{ }(n=k):}$
Now, assume true for fixed $k> 3$. We have a set of points: $$X := \{p_1,...,p_k\}\subseteq\mathbb{R}^2,$$ a surjective coloring function: $$f:X\to \{0,1\},$$ and there exists a line: $$\exists L:= L_{p_ip_j}:= \{p_i+(p_j-p_i)t\text{ }|\text{ }t\in \mathbb{R}\}\subseteq\mathbb{R}^2$$ such that: $$\forall x,y\in X\cap L:\text{ }f(x)=f(y).$$ Moreover, we have $X\cap L \neq X$ (by assumption all points of $X$ are not on the same line).
$\bullet\text{ }\underline{Step\text{ }Case\text{ }(n=k+1):}$
Given the previous case, let us append one more point to $X$. That is: $$\widetilde{X} := X\cup \{p_{k+1}\}\text{ }\text{ }\text{ and }\text{ }\text{ }Im(\widetilde{f}):= Im(f)\cup\{\widetilde{f}(p_{k+1})\}.$$ If $p_{k+1}\in X$ already, we are done by the previous case.
So let us assume $p_{k+1}\notin X$. Then using (a) in the induction hypothesis, we have $$\exists L:= L_{p_i,p_j}$$ going through two points $p_i,p_j\in X\subseteq \widetilde{X}$ and using (b) all points in $X\cap L$ have the same color. We need all points in $\widetilde{X}\cap L$ to have the same color to finish the $k+1$ case. The only question now is the state of $p_{k+1}$ (since we have as well not all points of $X$ are colinear implies not all points of $\widetilde{X}$ are colinear). There are four options: $\big(p_{k+1}\in L$ or $p_{k+1}\notin L\big)$ and $\big(\widetilde{f}(p_{k+1}) = 0$ or $1\big)$.
If $p_{k+1}\notin L$ then $\widetilde{X}\cap L=X\cap L$ and we're done.
Otherwise, with $p_{k+1}\in L$, success $\color{red}{\text{depends on our choice of extension}}$ for $\widetilde{f}$... [x]