Union of a sequence of path connected sets being path connected Theorem:
Let ${A_n}$ be a sequence of path connected subsets of a space $X$ such that for each integer $n\ge 1$, $A_n$ has at least one point in common with one the preceding sets $A_1, ..., A_{n-1}$. Then $\bigcup_{n=1}^\infty A_n$ is path connected.
I am trying to prove this theorem however, I'm having trouble constructing a proof because I don't know how to deal with the condition that $A_n$ has at least one point in common with one the preceding sets $A_1, ..., A_{n-1}$. How can I prove this theorem?
 A: Let $x, y \in \bigcup_{n=1}^\infty A_n$. Then by definition, $x\in A_n$ and $y\in A_m$ for some $m, n$. So it suffices to show that 
$$\bigcup_{n=1}^K A_n$$
is path connected for all $K$, given the condition. We can do induction on $K$. So it suffices to show that if $A_1, A_2 \subset X$ are two path connected subset in $X$ with $A_1\cap A_2 \neq \emptyset$, then $A_1\cup A_2$ is also path connected. Is that easier now?
A: To show that a space is path connected is to show that if $x,y$ are two points in that space, there is a path from $x$ to $y$.
If $x,y\in\bigcup_{n=1}^\infty A_n$ then for some index $j$ we have $x\in A_j$ and for some index $k$ we have $y\in A_k$.
The set $A_j$ has a point $x_\ell$ in common with $A_\ell$ for some $\ell<j$, and $A_k$ has a point $x_m$ in common with $A_m$ for some $m<k$.  So there's a path from $x$ to $x_\ell$ and there's a path from $y$ to $y_m$.  Then $A_\ell$ has a point in common with some point in some subset with a smaller index than $\ell$ and that one has a point in common with some subset with a still smaller index, and so on.  EXCEPT, that we cannot go on forever getting smaller numbers in the set $\{1,2,3,\ldots\}$.  So eventually we'll have a path from $x$ to some point in $A_1$.  And similarly a path from $y$ to some point in $A_1$.  And then we can draw a path connecting those two points in $A_1$.  Hence a path from $x$ to $y$.
(Of course, it's possible that we'll get a path from $x$ to $y$ without going as far as $A_1$, but if all else fails, that does it.)
