What is the simplest proof that for every cartesian square $A\times A$, which is a subset of $\Gamma$, the set $A$ is finite?
$$\Gamma = \left\{ ( x ; x) \hspace{.5em} | \hspace{.5em} x \in [ 0 ; 1] \right\} \cup \left\{ ( x ; 0) \hspace{.5em} | \hspace{.5em} x \in [ 0 ; 1] \right\} \cup \left\{ ( 0 ; x) \hspace{.5em} | \hspace{.5em} x \in [ 0 ; 1] \right\} .$$
The graph of $\Gamma$ follows:
Suppose that $A$ contains more than one non-zero point: $y,z$. Without loss of generality $y<z$. So we have that $(y,z)$ is not of the form $(y,y)$ nor $(0,y)$ nor $(y,0)$. Therefore $A\times A$ is not a subset of $\Gamma$.
So it's not only finite, it has at most one non-zero point, so at most two points.