You went astray in counting the number of unacceptable committees. To form an unacceptable committee, you must first choose the two enemies and then fill out the rest of the committee with any $2$ of the remaining $5$ people, so there are $\binom52=10$ unacceptable committees. Your final result will then be $35-10=25$, which is correct.
You can also calculate the number of acceptable committees directly. There are $\binom54=5$ committees that can be formed without including either of enemies. If you include one of the enemies, there are $2$ ways to choose which one to include and then $\binom53$ ways to choose the other $3$ committee members, for a total of $\binom54+2\binom53=2\cdot10=5+20=25$ possible committees.
Added: When you group the enemies as one entity and then choose $4$ of the resulting $6$, you’re making two mistakes. First, those $\binom64=15$ foursomes include the $\binom54$ committees that can be chosen from the $5$ non-enemies. Secondly, if the enemy pair is one of the $4$ that you choose, you’re also choosing three more people, so you’re making a $5$-person committee.