It seems if you take $B=\mathbb{R}^2$ and $B'$ the complement of the closure of $\{(x,\sin(\frac{1}{x}));x\in(0,\infty)\}$$\Big\{\big(x,\sin\big(\frac{1}{x}\big)\big);x\in(0,\infty)\Big\}$ this is a counterexample. (Added bonus: $B'$ is also simply connected)