[proofplan] We prove the equivalence-relation axioms directly from the definition of $\sim$. Reflexivity comes from the connectedness of singleton subsets, symmetry is immediate from the symmetric condition of lying in a common connected subset, and transitivity follows from the fact that the union of two connected subsets with nonempty intersection is connected. Finally, we compare the definition of the equivalence class $[x]_{\sim}$ with the definition of $C_x$ as the union of all connected subsets containing $x$. [/proofplan]

[step:Use connected singletons to prove reflexivity] Let $x \in X$. The singleton subset $\{x\} \subset X$, equipped with the [subspace topology](/page/Subspace%20Topology) inherited from $\tau$, is connected: if $\{x\}$ were written as the union of two disjoint subsets that are both open and closed in $\{x\}$, then one of the two subsets would have to contain $x$ and the other would be empty. Thus no separation of $\{x\}$ exists. Since $\{x\}$ is connected and contains both $x$ and $x$, the definition of $\sim$ gives $x \sim x$. Therefore $\sim$ is reflexive. [/step]

[step:Read symmetry directly from the definition] Let $x,y \in X$ and assume $x \sim y$. By definition of $\sim$, there exists a connected subset $A \subset X$ such that $x \in A$ and $y \in A$. The same subset $A$ also contains $y$ and $x$, so the definition gives $y \sim x$. Therefore $\sim$ is symmetric. [/step]

[step:Join overlapping connected subsets to prove transitivity]Let $x,y,z \in X$ and assume $x \sim y$ and $y \sim z$. By definition of $\sim$, there are connected subsets $A,B \subset X$ such that $x,y \in A$ and $y,z \in B$. Since $y \in A \cap B$, the intersection $A \cap B$ is nonempty. [claim:The union of two connected subsets with nonempty intersection is connected] Let $A,B \subset X$ be connected subsets in the subspace topology, and suppose $A \cap B \neq \varnothing$. Then $A \cup B$ is connected in the subspace topology. [/claim] [proof] Suppose, toward a contradiction, that $A \cup B$ is disconnected. Then there exist disjoint nonempty subsets $U,V \subset A \cup B$ that are open in the subspace topology on $A \cup B$ and satisfy \begin{align*} A \cup B = U \cup V. \end{align*} The sets $U \cap A$ and $V \cap A$ are open in the subspace topology on $A$, disjoint, and their union is $A$. Since $A$ is connected, one of them is empty. Hence $A \subset U$ or $A \subset V$. The same argument applied to $B$ gives $B \subset U$ or $B \subset V$. Choose a point $p \in A \cap B$. If $A \subset U$, then $p \in U$, and since $p \in B$, the alternative $B \subset V$ is impossible because $U \cap V = \varnothing$. Thus $B \subset U$. Similarly, if $A \subset V$, then $B \subset V$. Therefore $A \cup B$ is contained entirely in one of $U$ or $V$, contradicting that both $U$ and $V$ are nonempty and cover $A \cup B$. Hence $A \cup B$ is connected. [/proof] Applying the claim to the connected subsets $A$ and $B$, we get that $A \cup B$ is connected. Since $x \in A \subset A \cup B$ and $z \in B \subset A \cup B$, there exists a connected subset of $X$ containing both $x$ and $z$. Therefore $x \sim z$, and $\sim$ is transitive.[/step]

[guided]We need to prove that $x \sim z$ from the assumptions $x \sim y$ and $y \sim z$. Unpacking the definition of $\sim$ is the essential first move. The assumption $x \sim y$ gives a connected subset $A \subset X$ such that $x \in A$ and $y \in A$. The assumption $y \sim z$ gives a connected subset $B \subset X$ such that $y \in B$ and $z \in B$. The point $y$ is the bridge between the two connected sets: since $y \in A$ and $y \in B$, we have $y \in A \cap B$, so $A \cap B \neq \varnothing$. We now prove that this shared point forces $A \cup B$ to be connected. Suppose, for contradiction, that $A \cup B$ is disconnected in its subspace topology. Then there are disjoint nonempty subsets $U,V \subset A \cup B$, open in the subspace topology on $A \cup B$, such that \begin{align*} A \cup B = U \cup V. \end{align*} Intersect this separation with $A$. The subsets $U \cap A$ and $V \cap A$ are open in the subspace topology on $A$, they are disjoint, and their union is $A$. Since $A$ is connected, this cannot be a separation of $A$. Therefore one of $U \cap A$ and $V \cap A$ is empty, which means $A \subset U$ or $A \subset V$. The identical argument applies to $B$: the subsets $U \cap B$ and $V \cap B$ are disjoint, open in the subspace topology on $B$, and cover $B$. Since $B$ is connected, either $B \subset U$ or $B \subset V$. Now use the common point. Choose $p \in A \cap B$. If $A \subset U$, then $p \in U$. Because $p \in B$, the set $B$ cannot be contained in $V$, since that would put $p$ in both $U$ and $V$, contradicting $U \cap V = \varnothing$. Hence $B \subset U$, so $A \cup B \subset U$, contradicting the nonemptiness of $V$. The case $A \subset V$ is the same with $U$ and $V$ interchanged. Thus $A \cup B$ is connected. Since $A \cup B$ is connected and contains $x$ through $A$ and $z$ through $B$, the definition of $\sim$ gives $x \sim z$. This proves transitivity.[/guided]

[step:Identify each equivalence class with the connected component] Fix $x \in X$. The equivalence class of $x$ is \begin{align*} [x]_{\sim} = \{y \in X : x \sim y\}. \end{align*} First let $y \in [x]_{\sim}$. Then $x \sim y$, so there exists a connected subset $A \subset X$ with $x \in A$ and $y \in A$. By the definition of $C_x$ as the union of all connected subsets of $X$ containing $x$, every point of $A$ belongs to $C_x$. Hence $y \in C_x$. This proves $[x]_{\sim} \subset C_x$. Conversely, let $y \in C_x$. By definition of the union \begin{align*} C_x = \bigcup \{A \subset X : x \in A \text{ and } A \text{ is connected in the subspace topology}\}, \end{align*} there exists a connected subset $A \subset X$ such that $x \in A$ and $y \in A$. Therefore $x \sim y$, so $y \in [x]_{\sim}$. This proves $C_x \subset [x]_{\sim}$. Combining the two inclusions gives \begin{align*} [x]_{\sim} = C_x. \end{align*} Since $x \in X$ was arbitrary, this equality holds for every $x \in X$. [/step]