If $X$ is a topological space and $A\subset X$ is a connected subset of $X$ and $Β$ is an open subset of $X$, is it true that $A\cap B$ is connected in the subspace $B$? I tried to prove this using the classic definition of connectedness but I can't get around it.
I actually need this for the special case where $X=\mathbb{R}^2$ and $A$ is a proper, open and connected subset of $\mathbb{R}^2$ and $B=\mathbb{R}^2\setminus K$, where $K$ is a compact subset of $A$, so maybe it holds in this case, maybe employing the equivalent path-connectedness of domains in $\mathbb{R}^2$.
Any ideas?
$\textbf{Edit:}$ As pointed out, the first part gets a negative answer. What about the 2nd part though? What if $B=\mathbb{R}^2\setminus K$, where $K\subset A$ is a compact set and $A$ is open?
The answer is no. Consider the following subsets of $\mathbb R^2$
$$A=\{(x,y)\in\mathbb R^2\mid 2<x^2+y^2<4,\;y>-1\}$$$$B=\{(x,y)\in\mathbb R^2\mid 2<x^2+y^2<4,\;y<1\}$$
Note that both sets are connected and open, but their intersection isn't (the sets form horseshoe shapes and intersect at the ends of their respective horseshoe).