Let $A$ be a connected subspace of $X$. If $A\subseteq B\subseteq \overline{A}$, then $B$ is also connected.
My attempt:
Approach(1): Assume towards contradiction, $B$ is not connected. Then $\exists P,Q\in \mathcal{T}_B$ such that $P,Q\neq \phi$, $P\cap Q=\phi$, and $P\cup Q=B$. Let $P=B\cap R$ and $Q=B\cap S$, where $R,S\in \mathcal{T}_X$. Since $P,Q\neq \phi$, $\exists x\in P=B\cap R$ and $\exists y\in Q=B\cap S$, $x\neq y$, because $P\cap Q=\phi$. Since $x\in R$ and $y\in S$, we have $R\in \mathcal{N}_x$ and $S\in \mathcal{N}_y$. So $x,y\in B\subseteq \overline{A}$. By definition of closure, $A\cap R\neq \phi$ and $A\cap S\neq \phi$. Both $A\cap R$, $A\cap S\in \mathcal{T}_A$. Since $P\cap Q=B\cap (R\cap S)=\phi$, we have $(A\cap R)\cap (A\cap S)=A\cap (R\cap S)\subseteq B\cap (R\cap S)=\phi$, inclusion follows from the hypothesis $A\subseteq B$. Hence $A\cap R$ and $A\cap S$ is disjoint set. $P\cup Q =B\cap (R\cup S)=B$. Which implies $A\subseteq B\subseteq (R\cup S)$. So $(A\cap R)\cup (A\cap S)=A \cap (R\cup S)=A$. Hence union of $A\cap R$ and $A\cap S$ is $A$. Thus $A\cap R$ and $A\cap S$ form a separation of subspace $A$. Which contradicts our initial assumption that $A$ is connected. Is this proof correct?
Approach(2): Assume towards contradiction. Let $P$ and $Q$ be the separation of $B$. By lemma 23.2, $A\subseteq P$ or $A\subseteq Q$. WLOG, assume $A\subseteq P$. Since $P$ is closed in $B$, $(\overline{A})_B$(Closure in $B$)$\subseteq P$, by property of closure set. Note $(\overline{A})_B=\overline{A}\cap B$, $\overline{A}$ denote closure in $X$. Since $B\subseteq \overline{A}$, we have $\overline{A}\cap B=B\subseteq P$. We already know $P\subseteq B$. Thus $B=P$. So $P\cup Q=B\cup Q=B$ which implies $Q=\phi$. Thus we reach contradiction. This proof is slight variation of Munkres’ proof. Is this proof correct?
Approach(3): (Munkres’ Proof) Assume towards contradiction. Let $P$ and $Q$ be the separation of $B$. By lemma 23.2, $A\subseteq P$ or $A\subseteq Q$. WLOG, assume $A\subseteq P$. By exercise 6 section 18(a) (link: Exercise 6, Section 17 of Munkres’ Topology), $B\subseteq \overline{A} \subseteq \overline{P}$. Since $P$ is closed in $B$, $(\overline{P})_B=P=\overline{P} \cap B$. So $P\cap Q= (\overline{P}\cap Q)\cap B$. Since $(\overline{P}\cap Q)\subseteq Q\subseteq B$, we have $P\cap Q= (\overline{P}\cap Q)\cap B=\overline{P}\cap Q=\phi$. Thus $\overline{P}\cap Q=\phi$. $B\cap Q\subseteq \overline{A} \cap Q\subseteq \overline{P}\cap Q=\phi$. Which implies $B\cap Q=\phi \Rightarrow Q=\phi$. Which contradict our initial assumption. Is this explanation correct? Munkres didn’t explicitly go into the subspace details. In this approach, I tried to fill in the details.
At first glance Munkres’ proof looks fine, but it’s not the “complete” proof(IMO), I didn’t realise this thing at first.
Edit: Better version of approach(3) is the following: $B\subseteq \overline{A} \subseteq \overline{P}$. So $B\subseteq \overline{P}$. $B\subseteq \overline{P} \cap B=(\overline{P})_B=P$, since $P$ is closed in $B$. Thus $B\subseteq P$, which implies $Q=\phi$. We reach contradiction.
Best Answer
I'll give my favourite proof: let $f: B \to \{0,1\}$ be continuous (where the codomain is discrete). Then $f\restriction_A$ is continuous and by connectedness of $A$, $f[A]=\{i\}$ for some $i \in \{0,1\}$. As $f$ and the constant function with value $i$ coincide on the dense subset $A$ and $\{0,1\}$ is Hausdorff, we have that $f \equiv i$ and so $B$ is connected. Or use that by continuity $f[B]\subseteq f[\overline{A}] \subseteq \overline{f[A]}=\overline{\{i\}}=\{i\}$ instead, of you prefer.
This uses the standard alternative definition of connectedness of $X$ by saying $X$ is connected iff every continuous $f:X \to \{0,1\}$ must be constant. (Sketch of proof: if it were non-constant, then $f^{-1}[\{0\}$ and $f^{-1}[\{1\}$ would form a separation of $X$ and if $A \cup B$ is a non-trivial separation into two disjoint open sets, we can define $F \equiv 0$ on $A$ and $1$ on $B$ to find a non-constant $f$, as required). I like such proofs better than messing with separations.