The question see above. I thought there is a counter-example but not sure:
$A:=\{(x,\sin\frac{\pi}{x}):x\in(0,1]\}$, Then $A'=\{(0,y):y\in [-1,1]\}$. Take $\overline{A}$ and connect $(1,0)$, $(0,1)$ with some arbitrary Jordan curve. It's simply connected, and I think a homotopy contracting it to a point would be impossible due to the continuousness problem at $(0,y)\times 0$ of $H(\vec{x},t)$, but I don't know how to give a strict proof. Any help would be appreciated.