take an example
L1= (a^n b^n | where n >=1 ) //CFL
L2=(a^n b^n c^n| where n >=1 ) // NCFL
there intersection are not CFL
BUT if
L1= (a^n b^n | where n >=0 ) //CFL
L2=(a^n b^n c^n| where n >=0 ) // NCFL
then there intersection will be empty language which is regular which is CFL
hence we may contradict here
BUT we always need do focus on worst case
so this statement is FALSE