Connecting points with closed sets Let $A$ be a connected open set in an arbitrary topology. Let $x,y \in A$. Is there always a connected closed $B \subseteq A: x,y \in B$? If not, is it true under some reasonably weak additional assumptions (like, for instance, axioms of separation)?
 A: This is not a complete answer but shows that separation axioms are not enough, and, if the answer is to be positive, one must also assume the space is locally connected. 
Take $X$ to be the closed topologist's sine curve, that is the graph of $\sin\frac1x$ for $0<x\le1$ together with the closed vertical line segment between points $(0,-1)$ and $(0,1)$. Let $A=X\setminus\{(0,\pm1)\}$. Then $A$ is open in $X$ and connected. Let $x=(0,0)$ and $y=(\frac1\pi,0)$. A closed subset $B$ of $A$ will have to miss some neighborhoods of $(0,-1)$ and $(0,1)$, and hence cannot be connected, if it contains both $x$ and $y$. 
A: Consider the space $S =\{0,1\}$ with the topology $\{ \emptyset, \{0\}, S\}$. Let $A = \{0\}$ and $x=y=0$. Then $A$ is open and connected but does not contain any nonempty closed subset.
A: Assume $A$ is a proper non-empty subset of $X$ where $X$ is endowed with the topology $\{\emptyset,A,X\}$. Then clearly $A$ is connected, but the only closed subset is the empty set. Clearly $x,y\notin\emptyset$.
