Let \(A \subset X\) be a connected subspace.
Show that if \(B\subset X\) satisfies \(A\subseteq B \subseteq \bar{A}\), then \(B\) is connected.
Solution
1. Suppose \(B=U\sqcup V\) were a separation of \(B\).
We show one side must be empty.
2. Since \(A\) is connected and \(A\subseteq B\), all of \(A\) lies in one side, say \(A\subseteq U\).
The intersections \(A\cap U\) and \(A\cap V\) would otherwise separate \(A\).
3. No point of \(V\) can lie in \(\overline A\).
Because \(V\) is open in \(B\), for each \(v\in V\) there is an open set \(O\subset X\) with \(v\in O\) and \(O\cap B\subseteq V\). Since \(A\subseteq U\), this gives \(O\cap A=\varnothing\), so \(v\notin\overline A\).
4. But \(B\subseteq\overline A\), so \(V=\varnothing\), a contradiction. Thus \(B\) is connected.
Combine the hypothesis with 3.