I came across this definition
Let $U\subset S\subset C$. We say that $U$ is relatively open in $S$ if for every $z_0 \in U$, there is $r > 0$ such that $$D(z_0 ;r)\cap S\subset U$$:
Now the author doesnt specify if the subset symbol means that you do not count the trivial subsets ( the empty set and the set itself).
So then i came across another definition.
Definition :A set $S \subset C$ is called connected if the only relatively open and closed sets in S are the empty set and S.
Then i thought take the union of 3 disconnected open balls in C call it $K$.I know this set is disconnected so i want to prove it by the definition above.All i have to do is a find a relatively open closed subset in it except the trivials.SO there must be a closed subset call it $L$ of this set that is relatively open in it Because it is not connected.This means that for every $z_0$ in this closed set there exist an $r>0$ such that if i take the r-ball $\cap K\subset L$ .
Now i am trying to find a pictorial way for it.I drew 3 open circles.And i try to find a closed subset that for any element in it the $r$-ball is inside the closed set.there is no problem for the elements inside the closed set.But at the boundary of my closed subset cant find an $r$-ball without containing some elements that belong to $K$ and not in my closed set so the ball cant be a subset of my of my closed set.
Note that each one of the three balls is relatively open AND CLOSED in $K$, being the intersection of a closed ball of $\mathbb{C}$ with $K$. For instance, $B_1 = \overline{B_1} \cap K$, where the closure is meant in $\mathbb{C}$. So you have found a subset of $K$ (namely, $B_1$) which is relatively closed and open in $K$, hence $K$ is disconnected.
Your definition of relatively closed sets is equivalent (in $\mathbb{C}$ and in any other metric space, if with $D(z_0;r)$ you mean the open ball of radius $r$ around $z_0$) to the one I gave in a previous comment (you could try to show this), which is probably easier to visualize.