[proofplan] We prove both directions by translating between separations of the subspace $A$ and non-empty proper clopen subsets of $A$. If $A$ is disconnected, the subspace separation criterion gives a decomposition $A = U \cup V$ by non-empty disjoint relatively open sets; one side is then relatively closed because its complement in $A$ is the other side. Conversely, a non-empty proper clopen subset $B$ has a non-empty relatively open complement $A \setminus B$, and the two sets form a separation of $A$. [/proofplan] [step:Turn a separation of $A$ into a non-empty proper clopen subset] Assume that $A$ is disconnected. Since $A$ is equipped with the [subspace topology](/page/Subspace%20Topology), [citetheorem:8885] gives non-empty subsets $U,V \subset A$ that are open in $A$, disjoint, and satisfy $U \cup V = A$. Define $B := U$. Then $B \neq \varnothing$ because $U \neq \varnothing$. Also $B \subsetneq A$: indeed, $V \neq \varnothing$ and $U \cap V = \varnothing$, so any point of $V$ belongs to $A \setminus B$. It remains to check that $B$ is clopen in $A$. The set $B = U$ is open in $A$ by construction. Moreover, \begin{align*} A \setminus B = A \setminus U = V. \end{align*} Since $V$ is open in $A$, the complement of $B$ in $A$ is open in $A$, so $B$ is closed in $A$. Hence $B$ is clopen in $A$, non-empty, and proper. [guided] Assume that $A$ is disconnected. The relevant topology throughout is the subspace topology on $A$, not the ambient topology $\tau$ on $X$. Because the hypotheses of [citetheorem:8885] are exactly that $(X,\tau)$ is a [topological space](/page/Topological%20Space) and $A \subset X$ is considered with the subspace topology, we may apply it to obtain subsets $U,V \subset A$ such that $U$ and $V$ are open in $A$, both are non-empty, $U \cap V = \varnothing$, and \begin{align*} U \cup V = A. \end{align*} We choose one side of this separation and define $B := U$. This immediately gives $B \neq \varnothing$, since $U \neq \varnothing$. To see that $B$ is a proper subset of $A$, use the other side of the separation: since $V \neq \varnothing$, choose an element $v \in V$. Because $U \cap V = \varnothing$, this element satisfies $v \notin U = B$, while $v \in V \subset A$. Thus $A$ contains a point not in $B$, so $B \subsetneq A$. Now we verify the two parts of being clopen relative to $A$. First, $B = U$ is open in $A$ by the separation. Second, the complement of $B$ inside $A$ is exactly $V$: \begin{align*} A \setminus B = A \setminus U = V. \end{align*} The equality holds because $U$ and $V$ are disjoint and their union is all of $A$. Since $V$ is open in $A$, the complement of $B$ in $A$ is open in $A$. By the definition of relative closedness, this means $B$ is closed in $A$. Therefore $B$ is both open and closed in $A$, non-empty, and proper. [/guided] [/step] [step:Turn a non-empty proper clopen subset into a separation of $A$] Conversely, suppose there exists a subset $B \subset A$ such that $B$ is clopen in $A$, $B \neq \varnothing$, and $B \subsetneq A$. Define \begin{align*} C := A \setminus B. \end{align*} Since $B \subsetneq A$, the set $C$ is non-empty. Since $B$ is closed in $A$, its complement $C$ is open in $A$. Since $B$ is open in $A$ by clopenness, both $B$ and $C$ are open in $A$. The sets $B$ and $C$ are disjoint by definition of set difference, and they satisfy \begin{align*} B \cup C = B \cup (A \setminus B) = A. \end{align*} Thus $B$ and $C$ are non-empty disjoint subsets of $A$ that are open in $A$ and whose union is $A$. By [citetheorem:8885], the subspace $A$ is disconnected. [/step]