[proofplan] We treat $A$ as the [topological space](/page/Topological%20Space) $(A,\tau_A)$, where $\tau_A$ is the [subspace topology](/page/Subspace%20Topology) inherited from $(X,\tau)$. The assertion is exactly the separation characterization of disconnectedness for this particular topological space. The forward implication unpacks the definition of disconnectedness, and the reverse implication verifies that the given pair $(U,V)$ is a separation of $(A,\tau_A)$. [/proofplan]

[step:Regard $A$ as a topological space with its subspace topology] Let $\tau_A := \{A \cap O : O \in \tau\}$ denote the subspace topology on $A$. Throughout the proof, when a subset of $A$ is said to be open in $A$, this means that it belongs to $\tau_A$. A separation of the topological space $(A,\tau_A)$ means an ordered pair $(U,V)$ of subsets of $A$ such that $U \neq \varnothing$, $V \neq \varnothing$, $U,V \in \tau_A$, $U \cap V = \varnothing$, and $U \cup V = A$. By definition, $(A,\tau_A)$ is disconnected precisely when such a separation exists. [/step]

[step:Unpack disconnectedness to obtain two open pieces covering $A$]Assume that $(A,\tau_A)$ is disconnected. By the definition of disconnectedness recalled above, there exists a separation $(U,V)$ of $(A,\tau_A)$. Therefore $U,V \subset A$, the sets $U$ and $V$ are non-empty, $U,V \in \tau_A$, they are disjoint, and their union is all of $A$: $U \cap V = \varnothing$ and $U \cup V = A$. Thus the required subsets $U$ and $V$ exist.[/step]

[guided]Assume that the subspace $(A,\tau_A)$ is disconnected. The word “subspace” matters only through the topology $\tau_A$: all openness statements are now interpreted inside $A$, not necessarily inside $X$. By definition, disconnectedness of the topological space $(A,\tau_A)$ means that $(A,\tau_A)$ admits a separation. Thus there are subsets $U,V \subset A$ satisfying $U \neq \varnothing$, $V \neq \varnothing$, $U,V \in \tau_A$, $U \cap V = \varnothing$, and $U \cup V = A$. Since membership in $\tau_A$ is exactly openness in the subspace topology, $U$ and $V$ are open in $A$. These are precisely the two non-empty disjoint open subsets of $A$ whose union is $A$, so the forward implication follows.[/guided]

[step:Use the given open cover by disjoint non-empty sets as a separation] Conversely, suppose there exist subsets $U,V \subset A$ such that $U \neq \varnothing$, $V \neq \varnothing$, $U,V \in \tau_A$, $U \cap V = \varnothing$, and $U \cup V = A$. These five conditions say exactly that $(U,V)$ is a separation of the topological space $(A,\tau_A)$. Hence $(A,\tau_A)$ is disconnected by definition. [/step]