If each of $C~\&~D$ is an event then $C \cap D = \emptyset \; \Rightarrow \;P(C \cup D) = P(C) + P(D)$.
Consider this: $B = A \cup \left( {B\backslash A} \right)\;\& \;A \cap \left( {B\backslash A} \right) = \emptyset$.
So, $P(B) = P(A) + P(B\backslash A) \geqslant P(A)\text{, because }P(B\backslash A) \geqslant 0$