Question regarding an equivalent definition of connected sets. Consider the statement :
In a metric space $X$, a subset $E$ is disconnected iff $E\subset A\cup B$ for some non-empty disjoint open sets $A,B$ in $X$ such that $E\cap A\neq \phi$ and $E\cap B\neq \phi$.
Now if I replace $X$ to be a topological space rather than a metric space.Then I doubt that this would hold.The if part would hold obviously but the only if part may not hold.Because if $E$ is disconnected,then we can find $A,B$ open in $X$,non-empty such that $A\cap E$ and $B\cap E$ are also non-empty and $E=(A\cap E)\cup(B\cap E)\subset A\cup B$,but we cannot claim that there are $A,B$ which are disjoint also.
I just want to clear the doubt because I have yet not studied topology and not capable of coming with a counterexample.Is my claim correct?
I topological spaces,I think I have to be satisfied with,
In a topological space $X$, a subset $E$ is disconnected if $E\subset A\cup B$ for some non-empty disjoint open sets $A,B$ in $X$ such that $E\cap A\neq \phi$ and $E\cap B\neq \phi$.
 A: You are correct about the statement not holding for general topological spaces.
Consider the following space $X$ : it has $3$ points $a,b,c$ , and opens $\emptyset, X, \{a,b\}, \{c,b\}, \{b\}$. 
Then $\{a,c\}$ is disconnected : indeed the subspace topology is simply the discrete topology, so you can see $\{a\}\cup \{b\}$ is a witness to this fact. 
However, any two opens that cover this subspace must intersect (proof : inspect the possible covers !)
In metric spaces, a possible proof of the statement clearly uses the metric : let $A',B'$ be opens of $E$ such that $E= A'\sqcup B'$. Consider for each $x\in A'$ a real number $\epsilon_x >0$ such that $B(x,\epsilon_x)\cap E \subset A'$, and same for $y\in B'$ with $\delta_y$. 
Then consider $A= \bigcup_{x\in A'}B(x,\frac{\epsilon_x}{2})$, $B=\bigcup_{y\in B'}B(x,\frac{\delta_y}{2})$
If $z\in A\cap B$, then it is at distance $<\epsilon_x/2$ from some $x\in A'$ and $<\delta_y/2$ from some $y\in B'$. Assume wlog that $\epsilon_x <\delta_y$. Then by the triangle inequality, $x\in B(y,\delta_y)$, which contradicts $A'\cap B' = \emptyset$
You see here that the "ability to divde by $2$" is somewhat crucial. It's very likely though that this proof carries over to the context of uniform spaces if you know what those are (they're a bit more general than metric spaces, but you can still "divide by $2$")
A: For general spaces $A$ and $B$ need not be disjoint but must be disjoint on $E$, so
$E \cap A \cap B = \emptyset$.
Then we just have a disconnection by relatively open non-empty subsets. We then have for general spaces:

$E \subset X$ is disconected iff there exist open sets $A$ and $B$ such that $E \subseteq A \cup B, E \cap A \cap B = \emptyset, A \cap E \neq \emptyset, B \cap E \neq \emptyset$. 

