We can easily know that if random variables $X$ and $Y$ are independent, then:
$$Cov(X,Y) = E[(X-E[X])(E-E[Y])]$$ $$= E[XY]-E[X]E[Y]$$ $$= E[X]E[Y]-E[X]E[Y]$$ $$= 0$$
So, if random variables X and Y are independent, we get $Cov(X,Y)=0$.
However, if $Cov(X,Y)=0$, can we get a conclusion that $X$ and $Y$ independent? My intuition tells me that it may not be ture. How can I prove my intuition? How can I construct a bunch of counter examples (both in discrete cases and in continuous cases)?
No, we can’t. Consider for example, as X~Norm(0,1) and Y be square of Standard Normal. In that case you also receive Cov(X,Y)=0, since E(X)=0 and E(Y)=0. However X and Y are highly dependent, because the knowledge of value of X tells us a lot of things about Y.
However, it is true in case of Multivariate Normal. For instance, take X and be as Norm(0,1) and check if it is true. Take as Multi. Normal(X+Y,X-Y) and compute the Cov(X+Y,X-Y)=Var(X)-Cov(X,Y)+Cov(X,Y)-Var(Y)=0, since Var(X)=Var(Y)=1