Connectedness of topological subspace. I'm very confused about connectedness of a subspace. Let's say $A \subset X$ and that I want to show that $A$ is connected. So I suppose that $A$ is disconnected, can I say that there exists two closed subsets in X , say P, Q such that $A = P \cup Q$, $P \cap Q = \varnothing$? 
My question is about the phrase "closed subsets IN X". Is that right? Can I always say that?
 A: A set is disconnected if there are two disjoint open subsets which together cover the space. This implies that each is also closed, since each is the complement of the other.
If you want to show that $A$ is connected, you suppose to the contrary it is disconnected. This means that there are subsets of $A$ which are open, disjoint, and together cover $A$. Note that they are open in the topology inherited by $A$ as a subspace of $X$, rather than the topology of $X$ itself. 
A: You can say that if $A$ is disconnected 
that there are two disjoint non-empty closed sets $C$ and $D$ (closed in $A$!) such that 
$A = C \cup D$ and $C \cap D = \emptyset$.
In terms of closed subsets of $X$, we can write $C = \hat{C} \cap A$ and $C =\hat{D} \cap A$, where $\hat{C}, \hat{D}$ are closed in $X$. 
So we can conclude that $A \subseteq \hat{C} \cup \hat{D}$, $A \cap \hat{C} \cap\hat{D} = \emptyset$ ,$A \cap \hat{C} \neq \emptyset \neq A \cap \hat{D}$.
Such a pair $(\hat{C}, \hat{D})$ of closed (or equivalently open) subsets of $X$ can be called a separation of $A$ in $X$.
It's not the case that we can find non-empty closed subsets of $C$ and $D$ of $X$ such that $C \cap D = \emptyset$, $A = C \cup D$; for one, this would imply $A$ is closed in $X$, which it need not be. You have to use the more complicated one if you want to work with closed subsets in $X$.
