Theorem. (Zassenhaus) If $A$ and $C$ are subgroups of a group $G$ and if $B$ and $D$ are normal subgroups of $A$ and $C$, respectively, then there are isomorphisms $$ \frac{B\left(A \cap C\right)}{B\left(A \cap D\right)} \cong \frac{A \cap C}{\left(A \cap D\right)\left(B \cap C\right)} \cong \frac{D\left(A \cap C\right)}{D\left(B \cap C\right)} . $$
Proof. (...) Exercise 350 states that if $X, Y$, and $Z$ are subgroups of a group and $Y \subset X$, then $X \cap Y Z=Y(X \cap Z)$. This implies: (i) $$ \color{red}{B\left(A \cap D\right)\left(A \cap C\right)=B\left(A \cap C\right)} $$ $$ \text { (take } \left.X=A, Y=A \cap D, Z=C\right) \text {; } $$ (2) $$ \begin{aligned} & A \cap C \cap B\left(A \cap D\right)=\left(A \cap D\right)\left(B \cap C\right) ; \\ & \left(\text { take } X=A \cap C, Y=A \cap D, Z=B\right) . \end{aligned} $$
The proof is now complete.
My question: How does he get: $$ \color{red}{B\left(A \cap D\right)\left(A \cap C\right)=B\left(A \cap C\right)} $$ because all I get is by plugging in as he says $$ A \cap (A \cap D)C=(A \cap D)(A \cap C). $$
Since $D\leq C$, you have $C\leq(A\cap D)C\leq DC=C$. So $$A\cap(A\cap D)C = A\cap C.$$
Thus, $A\cap C = (A\cap D)(A\cap C)$, from your calculations.
Multiplying both sides on the left by $B$ we get $$B(A\cap C) = B(A\cap D)(A\cap C),$$ which is the equality you want.