How to finish? Connectedness If $A$ is a connected set and $\{A_i : i \in I\}$, $I$ an arbitrary set (can be countable or not) of connected sets.
How to show that if $A \cap A_i \neq \emptyset$ for all $i \in I$ then $A \cup (\cup_{i\in I} A_i)$ is connected?
I am trying to show that if $A \cap A_u \neq \emptyset~~ \forall i \in I$ then for all $i, j \in I$ $A_i\cap A_j \neq \emptyset.$ This enought to conclude the result. But I stuck here.
What I tried: Suppose that there are $i,j\in I$ such that $A_i \cap A_j  = \emptyset.$
Then since $A\cap A_i \neq \emptyset$ and $A\cap A_j \neq \emptyset$ it somehow induces me to think that is possible to obtain a split for $A$. I don't know how to proceed.
Thanks
 A: Hint that should make it clear how to proceed: The connected component that contains $A$, in $A\cup\big(\bigcup_{i\in I} A_i\big)$, contains a point in each $A_i$ by your non-empty intersection condition.
If some connected component contains a point in $A_i$, then can it happen that it does not contain all of $A_i$?
A: That $A_i \cap A_j \not= \emptyset$ is not always true. Consider for each $n \in \mathbb{Z}$: $A_n:=\{n\}\times \Bbb{R}$ as a subspace of $\Bbb{R}^2$ and let $A=\mathbb{R}\times \{0\}$. Then $A_n\cap A \not= \emptyset$ for each $n$, but $A_n \cap A_m  = \emptyset$ whenever $m\not=n$.
Let $X$ be a topological space. Then, show that every continuous function $f:X\rightarrow \{0,1\}$, where $\{0,1\}$ is endowed with the discrete topology,  is constant if and only if $X$ is connected.
Then, consider a family $(A_i)_I$ of connected sets in $Y$ and $A$ connected such that $A_i \cap A \not= \emptyset$ for each $i \in I$.
Now, $j$ be an element not in $I$ and let $J=I \cup\{j\}$ and $A_j=A$. Then, let $X= \bigcup_J A_j$ and $f:X\rightarrow \{0,1\}$ be continuous. It's sufficient to prove that $f$ is constant. Let $x_1$ and $x_2$ be elements of $X$. Then, there is an $i_1,i_2$ in $J$ such that $x_k\in A_{i_k}$ for $k=1,2$. 
Since there is an $y_k \in A_{i_k}\cap A$ for $k=1,2$, we have:
$$f(x_i)=f(y_i)$$ since $x_i,y_i \in A_{i_k}$ and $A_{i_k}$ connected. 
Then, since $A$ is connected we also have $f(y_1)=f(y_2).$
Thus $f(x_1)=f(y_1)=f(y_2)=f(x_2)$ and since $x_1$ and $x_2$ were arbitrary, we are done.
