I think this version of the topologist's sine curveThe Warsaw circle is compact and simply connected but there are obvious neighborhoods with no simply connected open refinement. This provides a counterexample for the question as originally worded but does not have connected complement.
TakeIt can be realized as the union of the following: $$\left\{(x,\sin\frac{1}{x}):0<x<1\right\}$$ $$0\times[-2,1]$$ $$[0,1]\times -2 $$ $$1\times[-2,\sin 1]$$
I think this version of the topologist's sine curve is compact and simply connected but there are obvious neighborhoods with no simply connected open refinement.
Take the union of the following: $$\left\{(x,\sin\frac{1}{x}):0<x<1\right\}$$ $$0\times[-2,1]$$ $$[0,1]\times -2 $$ $$1\times[-2,\sin 1]$$
The Warsaw circle is compact and simply connected but there are obvious neighborhoods with no simply connected open refinement. This provides a counterexample for the question as originally worded but does not have connected complement.
It can be realized as the union of the following: $$\left\{(x,\sin\frac{1}{x}):0<x<1\right\}$$ $$0\times[-2,1]$$ $$[0,1]\times -2 $$ $$1\times[-2,\sin 1]$$
I think this version of the topologist's sine curve is compact and simply connected but there are obvious neighborhoods with no simply connected open refinement.
Take the union of the following: $$\left\{(x,\sin\frac{1}{x}):0<x<1\right\}$$ $$0\times[-2,1]$$ $$[0,1]\times -2 $$ $$1\times[-2,\sin 1]$$
