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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.

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 $B$.

To make sense of your question, especially in light of the other question linked to by David, we should assume that you are speaking of subsets of the plane or some other ambient space, and that you don't mean just that $C$ is relatively closed in $B$, (or we could take $C=B$ and be done with it).

With this understanding, the answer to your first question is no. Let $A$ be a sequence on the $x$ axis converging to the origin in the plane. 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 there is no connected closed subset of $B$ containing $A$.

We can extend this to a counterexample for your second question as well, about path-connectedness, if 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 $C\subset B$ containing $A$ will have to contain all those lines and hence all of $B$ and hence not be closed in the plane.

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.

The answer to your first question is no. Let $A$ be a sequence on the $x$ axis converging to the origin in the plane. 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 there is no connected closed subset of $B$ containing $A$.

To make sense of your question, especially in light of the other question linked to by David, we should assume that you are speaking of subsets of the plane or some other ambient space, and that you don't mean just that $C$ is relatively closed in $B$, (or we could take $C=B$ and be done with it).

With this understanding, the answer to your first question is no. Let $A$ be a sequence on the $x$ axis converging to the origin in the plane. 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 there is no connected closed subset of $B$ containing $A$.

We can extend this to a counterexample for your second question as well, about path-connectedness, if 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 $C\subset B$ containing $A$ will have to contain all those lines and hence all of $B$ and hence not be closed in the plane.