Theorem 24.3 in Munkres's Topology I'm having a bit of a trouble understanding the proof that if $A$ and $B$ are subspaces of $X$ and $A \subseteq B \subseteq \operatorname{Clo}(A)$, then if $A$ is connected so is $B$ in Munkres's Topology. The proof he gave was: if $B$ has a seperation $C \cup D$ then without the lost of generality, we can assume $A$ is contained in $C$, but that means $\operatorname{Clo}(A) \subseteq \operatorname{Clo}(C)$. Since $\operatorname{Clo}(C)$ is disjoint from $D$, that contradicts the fact that $B$ contains $D$.
I think the closures are taken with respect to $X$, because then we can conclude that $B$ is a subset of $\operatorname{Clo}(C)$, but then I'm having a hard time figuring out how $\operatorname{Clo}(C)$ (taken with respect to $X$) is disjoint from $D$.
 A: Since $D$ is open in $B$ it follows that $D=D'\cap B$ where $D'$ is open in $X.$ Since $C\subseteq B$ we find that 
$$C\cap D'=C\cap (D'\cap B)=C\cap D=\emptyset.$$
Thus $D'$ and $C$ are disjoint.
The set $\text{Clo}(C)$ is the smallest closed set containing $C.$ The set $X\backslash D'$ is a closed set containing $C,$ so $\text{Clo}(C)\subseteq X\backslash D'.$ Since $D\subseteq D'$ we find $D\cap X\backslash D'=\emptyset,$ therefore $D\cap\text{Clo}(C)=\emptyset.$
A: $A$ is a dense subset of the space $B$ because $Cl_B(A)=B\cap  Cl_X(A)=B. $
If $A$ is a dense connected subspace of a space $B$ then $B$ is connected.
Proof: Let $U,V$ be disjoint open subsets of $B$ with $U\cup V=B.$ We want to show that at least one of $U,V$ is empty. 
Now $(U\cap A),(U\cap V)$ are disjoint open subsets of $A$ and their union is $A.$ Since $A$ is connected, at least one of $(U\cap A),(V\cap A)$ is empty. WLOG let $$U\cap A=\emptyset.$$
In any space an open set $O$ which is disjoint from a dense set $D$ must be the empty set (otherwise $D$ would not be dense).
And in the space $B$ the open set $U$ is disjoint from the dense subset $A,$ so $U=\emptyset.$
