= Solution
If $A$ is disconnected, a separation writes it as $A=A_1\cup A_2$ with $A_1,A_2$ disjoint, nonempty and open in its <subspace topology>. By definition there are ambient <open sets> $U,V$ with $A_1=U\cap A$ and $A_2=V\cap A$. These give exactly the four conditions stated.
Conversely, those conditions make $U\cap A,V\cap A$ two nonempty disjoint relatively <open sets> whose union is $A$. They are therefore a separation. This proves the equivalence. \b[The ambient sets may overlap outside $A$], which is why the condition here is $U\cap V\cap A=\varnothing$.
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