Why does the definition of (dis-) connectedness use open sets? My course on metric spaces uses the following definition for connectedness:

$(X,d)$ is disconnected if there exist open $U,V \subset X  $ such that $ X = U \cup V; \ U,V \neq \emptyset $ and $U \cap V = \emptyset$. $(X,d)$ is connected if it is not disconnected.

I started playing a bit with this definition and wondered how necessary the requirement of openness on $U$ and $V$ is. That is would we get an absurd definition if we allowed one set to be closed/clopen?
 A: Open disjoint sets is just one way for a space to be disconnected.
Exercise.
Prove if K,L are not empty, disjoint, closed subsets of S,
and K $\cup$ L = S, then S is disconnected.  
A: The original definition for disconnectedness is that there exists subsets $A,B$ of $X$ such that $X=A\cup B$ and $\emptyset \neq A \neq X$, $\emptyset \neq B \neq X$ ($A$ and $B$ are proper, non-empty subsets) and $A$ and $B$ are separated: $\overline{A} \cap B = \emptyset = A \cap \overline{B}$: no point can be in $A$ and close to $B$ too (or vice versa in $B$ and close to $A$), a stronger version of being merely disjoint. 
The basic idea is that $X$ has two pieces that are "apart from each other". 
Now, in the above situation, as $X=A \cup B$ we in fact have that $A$ and $B$ are both closed in $X$: no point of $\overline{A}$ can be in $B$, but it has to be in $A$ or $B$ so must be in $A$ and $A$ is thus closed, or formally 
$$A \subseteq \overline{A} \subseteq X\setminus B \subseteq A \text{ so } \overline{A}=A$$ and likewise for $B$.
But as $A$ and $B$ are each other's complement they are also both open in $X$.
So in a disconnecting partition we can ask for either two separated sets, or two disjoint open sets (which are automatically separated) or two disjoint closed sets (also automatically separated). So in a way it doesn't matter, a text has to choose one definition and stick to it. 
