Explanation of proof of why connectedness is not conjunctively local of any order $k$
The predicate $\varphi_0$ depends on at most $k$ points. There are $k+1$ middle squares. So $\varphi_0$ cannot depend on all of them. That is, there is a middle square that $\varphi_0$ does not depend on. Therefore, if we add that square, the value of $\varphi_0$ doesn't change, and so $\psi$ is still 0, even though the figure is now connected.