So far my proof goes like this. Let $(S, d)$ be a connected subspace of $(X, d)$. By contradiction, suppose $\overline S$ is not connected. Then, we can write $\overline S = A \cup B$, where $A, B$ are non-empty, disjoint, open (in $\overline S$) subsets of $\overline S$. Moreover, since $S \subset \overline S$, then $S = (S \cap A) \cup (S \cap B)$. Take $x \in S$, and assume without loss of generality that $x \in A$. Then it follows that $S\cap A \neq \varnothing$. How can I show that $S \cap B \neq \varnothing$ to get a contradiction?
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1$\begingroup$ Pick a point in $B$, and then consider a small open $U$ (in $\bar S$) around it. What can you say about $U$? $\endgroup$– NivenMay 30, 2021 at 23:23
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$\begingroup$ @Niven: $U \subset B \subset \overline S$? $\endgroup$– Rob32409May 30, 2021 at 23:34
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1$\begingroup$ Right, that's half of it, that $U\subset B$. You also know one more thing, using that $U$ is a neighborhood of a point in the closure. $\endgroup$– NivenMay 30, 2021 at 23:37
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$\begingroup$ @Niven: I think what follows is to say that $U \cap S \neq \varnothing$, so $B \cap S\neq \varnothing$, but how to show that $U \cap S \neq \varnothing$? $\endgroup$– Rob32409May 31, 2021 at 0:23
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2 Answers
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I am not sure how to hint more in the comments without giving away the answer, so I guess I should respond with an answer. Also, I should apologize that I made things more complicated than necessary in the comments; in particular, in the end, there was no real reason to introduce another small open $U$. I'm not sure why I thought there was. Anyways...
You can use the following characterization of closures: a point $p\in X$ is in the closure $\bar S$ of $S$ iff every neighborhood of $p$ intersects $S$ (otherwise, if $V$ is a neighborhood of $p$ disjoint from $S$, then $\bar S\cap(X\setminus V)$ would be a smaller closed set containing $S$, a contradiction).
In the present case, we pick $p\in B\subset\bar S$. Since $B$ is open, it is an open neighborhood of a point of the closure, so we must have $B\cap S\neq\emptyset$.
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Suppose $\bar{S}$ is not connected. Then let $A_1, B_1$ be closed in $(X, d)$ such that \begin{align} \bar{S}=(A_1 \cap \bar{S}) \cup (B_1 \cap \bar{S})=A \cup B \end{align} where $A_1 \cap B_1=\phi$ (also $A \cap B =\phi$). Now $S \cap A$ and $S \cap B$ are disjoint closed sets in $S$ such that \begin{align} S=(S \cap A)\cup(S \cap B) \end{align} Since $S$ is given connected we must've either $S \cap A =\phi$ or $S \cap B =\phi$. Without loss of generality let $S \cap A =\phi$, then $S \subseteq B$ implies $\bar{S} \subseteq \bar{B} \subseteq B_1 \cap \bar{S}=B$. Consequently, $A=\phi$, a contradiction. Thus $\bar{S}$ must be connected.
