We need to make the additional assumption that the graph G has at least two vertices. Otherwise a graph with a single vertex and zero edges is technically connected and has no cycles but does not have any vertices with degree 1.
Prove the equality of the two sets below by showing that every element of the set on the left hand side of the equality is also an element of the set on the right hand side, and vice versa. (I.e., do not prove this using algebraic identities.)
(B − A) ∪ (C − A) = (B ∪ C) − A