That could work as well. However, since the order in which the pairs of people are selected does not matter, we need to divide by $2!$ to get rid of repetitions. You could have chosen $A$ to be alone, $B$ and $C$ to share the first room $R_1$ (out of $4$ available), and $D$ and $E$ to share the second room $R_2$ (out of $3$ available), or you could have chosen $A$ to be alone, $D$ and $E$ to share the first room $R_2$ (out of $4$ available), and $B$ and $C$ to share the second room $R_1$ (out of $3$ available).

So, choosing two rooms in no particular order, and the two pairs of guests to go in each is the same as choosing the two pairs in no particular order, and then choose the two rooms: $$\binom{4}{2} \cdot \binom{4}{2}\binom{2}{2} = \dfrac{\displaystyle \binom{4}{2}\binom{2}{2}}{2!}\cdot 4 \cdot 3.$$