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Let $(X,d)$ be a metric space, show that if $A \subset X$ is connected, $B \subset X$ with $A \subset B \subset cl(A)$ then B is connected.

My approach:

Let's assume that B is not connnected, then by definition there exist two open, disjunct subsets $U, V \subset X$ such that $ B \subset U \cup V, B \cap V \neq \emptyset, B \cap U \neq \emptyset$. Let $b_1 \in B \cap U, b_2 \in B \cap V$. Then since $B \subset cl(A) \Rightarrow \forall \epsilon \gt 0: K(b_1, \epsilon) \cap A \neq \emptyset$ and $\forall \epsilon \gt 0: K(b_2, \epsilon) \cap A \neq \emptyset$

Now I'm not sure whether the next step is correct, to me it seems as if since $K(b_1, \epsilon) \cap A \neq \emptyset$ is true for all $\epsilon \gt 0$, there has to be an element $a_1 \in A : a_1 = b_1$ and the same logic would apply to b_2.

Then there would exist $U, V \subset X : A \subset U \cup V, A \cap V \neq \emptyset, A \cap U \neq \emptyset$, therefore A is not connected.

Is that correct?

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@Gregory is correct that this hypothesis is true for any topological space. But I think the construction of his problem is a little different, for to use his terminology, you are trying to show that $Z \subset X$ is connected given connectivity of $X$, not the other way around. I offer a proof of this statement here:

Keeping with the structure of the problem at hand, if $(X,d)$ is a metric space, then your proof veers off course a little early on. For if $B = C \cup D$ is a separation of $B$ and we know $A \subset B$, then the set $A$ must lie entirely in $C$ or $D$. Now pick one of the subsets $C$ or $D$ such that $A$ is contained entirely. Without loss of generality, one can assume $A \subset D$.

The next trick involves showing that $C$ and $\overline{D}$ are disjoint, which would imply $\overline{A} \subset \overline{D}$. But this is certainly the case; say we take a point $x \in \overline{D}$. Using metrizability of $X$, we then take an $\epsilon$-neighborhood of $x$; but we know this cannot intersect $C$ for arbitrary $\epsilon$ because the two sets are disjoint, so $x \notin \overline{C}$. So we have shown $\overline{A} \subset \overline{D}$, but since $B \subset \overline{A}$, this means that $C$ is empty, which contradicts that $C \cup D$ is a separation of $B$.

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You won't be able to show $b_1$ is in $A$, it may not be. The point though is that $U\cap A$ and $V\cap A$ partition $A$ and therefore one of them must be the whole of $A$ and the other must be empty. But if one of them can be empty and has non-empty intersection with $B$ then $A$ can't be dense in $B$, which it is by assumption.

Anyway, I think this is true in general, not just in a metric space. If $X$ is a topological space and $Z\subseteq X$, where $Z$ is connected and dense in $X$, then $X$ is connected.

Proof: Let $A$ be a non-empty subset of $X$ which is both open and closed. Since $Z$ is dense in $X$ we know that $Z$ must intersect every nonempty open subset of $X$, and therefore $A\cap Z$ is non-empty. Now $A\cap Z$ is both open and closed in $Z$, and since $Z$ is connected we deduce that $A\cap Z=Z$, i.e., $Z\subseteq A$. Therefore $X=\overline{Z}\subseteq\overline{A}=A$, giving $X=A$ as required.

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