Closed connected subset of a connected set Let $A$ be a closed set and let $B$ be a connected set such that $A \subset B$.
Does there always exist a closed connected subset $C$ of $B$ that contains $A$?
What if $B$ is path connected, is there always a path-connected $C$? A connected $C$?
 A: The answer to your first question is no. Let $A$ be a sequence on the $x$ axis converging to the origin in the real plane $\mathbb{R}^2$, which is the background topological space for these examples. Let $B$ be a cone over the sequence, connecting each point in the sequence to the point $(0,1)$, together with the origin, but not connecting the origin to $(0,1)$. Then $B$ is connected, but not closed, and there is no connected closed set $C$ contained in $B$ and containing $A$, since any connected set $C$ contained in $B$ and containing $A$ would have to contain all the lines and thus be all of $B$, which is not closed in the plane. 
The example can be extended to an answer for your path-connected question as well. We simply add a roundabout path from the origin to $(0,1)$. That is, $A$ is the sequence $(\frac{1}{n},0)$ plus $(0,0)$ in the real plane. The set $B$ contains lines from every $(\frac{1}{n},0)$ to $(0,1)$, and $B$ also includes a circular curve connecting $(0,0)$ to $(1,0)$ through the left half-plane. So $A$ is closed, $B$ is path-connected, but any closed path-connected set $C$ with $A\subset C\subset B$ will have to contain all those lines and hence $C=B$, but this is not closed in the plane. 
Similarly, for your last question, any closed connected $C$ with $A\subset C\subset B$ for the same example will have to contain all the lines from the points on the sequence to $(0,1)$, and hence not be closed (since it will miss the line joining the origin to $(0,1)$, which is not in $B$. 
