Hi I just want to find out if the following is a acceptable proof for the proposition:
"Consider metric space $(X,d)$. If $A \subset X$ is connected and $A \subset B \subset \bar{A}$ then $B$ is connected."
Proof:
Assume that $A \subset X$ is connected and $B$ is not connected. There exists disjoint open sets (in $(B,d)$) $V_{1}$ and $V_{2}$ such that $B = V_{1} \cup V_{2}$. Since $A \subset B$, it follows that $A \cap V_{1}$ and $A \cap V_{2}$ are open in $(A,d)$. Also then $A = (A \cap V_{1}) \cup (A \cap V_{2})$. Thus since $A $ is assumed connected, either $A \cap V_{1} = \emptyset$, $A \cap V_{2} = \emptyset$, $A \cap V_{1} = A$ or $A \cap V_{2} = A$.
W.L.O.G assume that $A \cap V_{1} = \emptyset$, then there exists an $x \in B \subset \bar{A}$ and $\epsilon > 0$ such that $B(x,\epsilon) \cap A = \emptyset$. This contradicts $x \in \bar{A}$. Assume now W.L.O.G that $A \cap V_{1} = A$, then $A \subset V_{1}$ and $A \cap V_{2} = \emptyset$. If we take $x \in V_{2} \subset B \subset \bar{A}$ then again we find $\epsilon > 0$ such that $B(x,\epsilon) \cap A =\emptyset$. This contradicts $x \in \bar{A}$. Therefore there cannot exist disjoint open sets $V_{1}$ and $V_{2}$ in $B$ such that $B = V_{1} \cup V_{2}$. Hence, $(B,d)$ cannot be disconnected and is therefore connected.
$\square$
Best Answer
I would add the word "non-empty" since you want to have a contradiction to the assumption that both sets are inhabited, and you later use that you can find an element in the "other" $V_i$.
Here it looks as if you distinguish four cases. However, $A\cap V_1=\emptyset$ is equivalent to $A\cap V_2=A$ (and $A\subseteq V_2$). So you have only two cases, and of these it suffices to consider the case $A\cap V_1=\emptyset$, since the situation is completely symmetric in $V_1$ and $V_2$ (That's what is expressed by the W.L.O.G.).
Otherwise, the proof is fine. You take an $x\in V_2$ and conclude by the openness of $V_2$ that the relative neighborhood $B(\varepsilon,x)\cap B$ is contained in $V_2$ and thus disjoint from $A$, contradicting $x\in\overline A$. At least that's what it looks like, as you didn't include all the details.