Almost, but not quite. To contradict the statement
cl(A\capB)\supseteq[cl A \cap cl B]
you must exhibit an element of [cl A \cap cl B] that is not in cl(A\capB). As you noted, if A=(0,1) and B=(1,2) such an element is 1, for
1\in[cl A \cap cl B] but 1\notincl(A\capB)=\phi