I want to show that the ordered square is connected but not path connected.
It's connected because it's a linear continuum. But its not path connected. This is where I'm stuck. So I took two points $(x_0,y_0)$ and $(x_1,y_1)$. The lines are the open neighborhood $U$ that contains both of those points.
Just by looking at the picture below, I'm stuck as to why its not path connected.
Sorry if the picture is poorly drawn, but I just learned how to use GIMP to sketch stuff out.