Question about definition of connectedness. While reading some books, I often find the following argument: 
Consider $A\subset X$, when $X$ is a topological space.
If $A$ is a disconnected set, then there exist some open sets, say $U,V \in \tau_X$ disjoint and non empty such that $A = U \cup V$.
But this often confuses me, because the part when it says "open sets of $\tau_X$" because, for example, $[0,1] \cup [5,6]$ is disconnected and it can't be written as union of two open sets -of X!- which are disjoint and non empty.
The definition in some books of disconnected set $A$ says that $A$ is disconnected if it is disconnected with the subspace topology. 
But then, I don't understand why some authors take open sets of the big space, say $X$.
Can someone help me to clarify this?
 A: 
Definition of Separation of a Topological Space. Let $(X,\tau_X)$ be a topological space and $Y\subseteq X$. Let the subspace topology on $Y$ be denoted as $\tau_Y$. A separation of $Y$ is a pair of disjoint non-empty open (open in $Y$, not $X$) subsets of $Y$ whose union is $Y$.
Definition of Disconnected Topological Space. A topological space $(X,\tau_X)$ is said to be disconnected if there exists a separation of $X$.

So when trying to prove that $[0,1]\cup [5,6]$ is disconnected we shouldn't be trying to write it as union of two sets open in $\mathbb{R}$, rather we should try to write it as an union of sets open in $[0,1]\cup [5,6]$ under the subspace topology induced by the topology on $\mathbb{R}$ under consideration.
A: Here is a way to clarify this..

Definition. A topological space $(X,\tau)$ is disconnected if there exist $U,V\in\tau\setminus\{\varnothing\}$ such that $U\cap V=\varnothing$ and $U\cup V = X$. Otherwise, $(X,\tau)$ is connected.
We say that a subset $Y\subseteq X$ is connected if it is connected as a topological subspace.

Proposition. Let $(X,\tau)$ be a topological space and $Y\subseteq X$.
We have that $Y$ is disconnected if and only if there exist $U,V\in \tau\setminus\{\varnothing\}$ such that
(1) $U\cap Y\neq\varnothing$;
(2) $V\cap Y\neq\varnothing$;
(3) $U\cap V\cap Y=\varnothing$; and
(4) $Y\subseteq U\cup V$.
Proof. (ONLY IF) Assume $Y$ is disconnected, then there exist $U_Y, V_Y\in\tau_Y\setminus\{\varnothing\}$ such that $U_Y \cap V_Y = \varnothing$ and $U_Y\cup V_Y = Y$.
Thus there exist $U,V\in\tau\setminus\{\varnothing\}$ such that $U_Y=U\cap Y$ and $V_Y=V\cap Y$.
Such $U$ and $V$ satisfy conditions (1)-(4).
(IF) Assume there exist such $U$ and $V$ in $\tau$ satisfying (1)-(4).
Let $U_Y=U\cap Y$ and $V_Y=V\cap Y$, both in $\tau_Y\setminus\{\varnothing\}$ (conditions (1) and (2) tell us that both $U_Y$ and $V_Y$ are nonempty).
Condition (3) tell us that $U_Y\cap V_Y= \varnothing$ and, finally, condition (4) says that $U_Y\cup V_Y=Y$.
Therefore, $Y$ is disconnected. $\square$
A: The correct definition of $A$ being a disconnected subset of $X$ is that $A$ is disconnected as a space in its own right, i.e. having the subspace topology.
If $A \subseteq X$ is a disconnected subset, we can translate this to a statement about open sets of $X$ if we like:

$A \subseteq X$ is disconnected iff there exist $U,V$, open sets in $X$, such that 
  
  
*
  
*$U \cap A \neq \emptyset$
  
*$V \cap A \neq \emptyset$
  
*$A \subseteq U \cup V$
  
*$A \cap U \cap V = \emptyset$
  

This corresponds to $\{U \cap A$, $V \cap A\}$ being a disconnection of $A$ in its subspace topology.
