If A and B are path connected, then AUB is path connected Let A and B be subsets of a topological space X such that $A \cap cl(B)\neq \emptyset$.  Prove or disprove the following statement:
If A and B are path-connected, then $A\cup B$ is path-connected.
Intuitively I think this is false since although A contains a limit point of B, it doesn't seem necessary for a path to exist between a point in B and $cl(B)$.  But I can't come up with a counterexample.
Can anyone help please?
Thank you
 A: Counter-example: The Topologists' Sine Curve, a subspace of  $\mathbb R^2,$ is $$X= A\cup B$$ $$ \text {where }\quad A=\{0\}\times [-1,1]$$ $$ \text { and }\quad  B=\{(x,\sin  (1/x)): x\in (0,1)\}.$$ We have $A=Cl_X(B).$ And clearly $A$ and $B$ are path-connected subspaces of $X$. But $X$ is not path-connected.
Notation: $B((0,v),d)=X\cap \{(x,y): |x|<d\land |y-v|<d\}.$ 
The idea of the proof is that for $(0,v)\in A,$ if  $d\in (0,1)$ and  $S\subset B((0,v),d)$ with $A\cap S\ne \emptyset \ne S\cap B,$ then $S$ is disconnected. 
Suppose $f:[0,1]\to X$ is continuous with $f(0)\in A$ and $f(1)\in B.$  Let $$s_0=\sup \{s\in [0,1]: A\supset \{f(t):t\in [0,s]\}\;\}.$$ Then $f(s_0)\in A$ because $f$ is continuous and $Cl_X(A)=A.$ And $s_0<1$ because $f(1)\in B.$
There exists $t\in (0,1-s_0)$ such that $$X\cap B(f(s_0),1/2)\supset S=\{f(s):s\in T\},$$ $$\text { where }\quad  T=[0,1]\cap (s_0-t,s_0+t).$$ But $S\cap A\ne \emptyset \ne S\cap B .$ (That is, $f(s_0)\in S\cap A,$ and $S\cap B\ne \emptyset$ by def'n of $s_0$). So $S$ is disconnected. But $S$ is the continuous image of the connected set $T$, a contradiction.  
Remark: $X$ is a favorite example of a connected space that is not path-connected. The connectness of $X$ follows from the fact that $B$ is a dense open connected subset of $X.$
