Covering consisting of connected subsets Let $X$ a topological space and suposse there exists a covering $\mathcal{A}$ of $X$ consisting of connected subsets of $X$ such that if $A,B \in \mathcal{A}$ then there is a finite subcollection $\{ A_1,A_2, \cdots, A_n \}$ of $\mathcal{A}$ such that $A=A_1$, $B=A_n$ and $A_{i} \cap A_{i+1} \neq \emptyset$ for each $i=1,2, \cdots, n-1$. And I have to prove that $X$ is connected. \
I only supposed that $X$ is not connected, then exists $U$ and $V$ open in $X$ such that $X= U  \cup V$ and $U \cap V = \emptyset$. Can anybody help me?
 A: Suppose indeed that $X= U \cup V$ where both are open, disjoint and non-empty.
Pick $p \in U$ and $q \in V$ and also some $A,B \in \mathcal{A}$ such that $p \in A$, $q \in B$ (which can be done as $\mathcal{A}$ is a cover by assumption).
Now apply the one thing we know: there are $A_1= A, A_2,A_3,\ldots, A_n = B$ all in $\mathcal{A}$ such that  $A_i \cap A_{i+1} \neq \emptyset$ for all $i=1,\ldots n-1$.
The other thing we know: all $A_i$ are connected, so as $p \in A_1 \cap U$ we know that $A_1 \subseteq U$ (or $\{U,V\}$ would be induce a relative disconnection of $A_1$ etc.) and as $A_2 \cap A_1 \neq \emptyset$, we know that $A_2 \cap U \neq \emptyset$ and so for the same reason $A_2 \subseteq U$, and so on until we conclude (in finitely many steps) that $U_n = B \subseteq U$ as well. But then we have a contradiction as $ q \in B \cap V \neq \emptyset$ and so $U$ and $V$ intersect, which was not the case. This proves that $X$ must be connected.
A: *

*In any  space $X$, if $C,D$ are connected subspaces of $X$ with $C\cap D\ne \phi$ then $C\cup D$ is connected.


*If $A_1,..,A_n\in \mathcal A$ with $A_j\cap A_{j+1}\ne \phi$ whenever $1\le j<n$ then $\cup_{j=1}^n A_j$ is connected. Proof: Case $n=1$ is trivial. If Case $n$ is true then Case $n+1$ is true by 1. with $C=\cup_{j=1}^nA_j$ and $D=A_{n+1}.$


*If $x,y\in X$ then there exist $A,B\in \mathcal A$ with $x\in A$ and $y\in B$ so by 2. there exist $A_1,...,A_n\in \mathcal A$ such that $x,y$ belong to the $connected$ subspace $\cup_{j=1}^n A_j.$


*If any $x,y \in X$  belong to some connected $S\subseteq X$ then $X$ is connected. Proof: Suppose $U,V$ are disjoint open subsets of $X$ with $ U\ne \phi$ and $U\cup V=X.$ Take $x\in U$. For any $y\in X,$ take a connected $S_y\subseteq X$ with  $\{x,y\}\subseteq S_y.$ Then $U_y'=U\cap S_y,\,V'_y=V\cap S_y$ are disjoint open subsets of the space $S_y,$ with $U'_y\cup V'_y=S_y$ and $U'_y\ne \phi $... (as $x\in U'_y\,$)... so $U_y'=S_y,$ as $S_y$ is connected. So $y\in U'_y\subseteq U$ for any $y\in X,$ so $U=X.$
Remark: We can also prove the first sentence of 4. as  follows: If $X$ is disconnected, take $x\in U$ and $y\in V$ where $U,V$ are disjoint open subsets of $X$ with $U\cup V=X.$ If $\{x,y\}\subseteq S\subseteq X$ then $\{S\cap U,\,S\cap V\}$ is a disconnection of $S,$  i.e. $S$ is not connected.
