Prove line in same plane as triangle that intersects interior, intersects a side

Please let me know if the following proof is correct or not. If not a hint would be appreciated.

Given $\triangle ABC$ and line l lying in the same plane as $\triangle ABC$ . Prove that if l intersects the INT( $\triangle ABC$ ) then it intersects at least one of the sides.
Proof: Let D and E be points on l and D,E $\in INT(\triangle ABC).$ Since D,E $\in INT(\triangle ABC)$ they are on the same side of $\overleftrightarrow{AC}$ as B. There are seven possibilities:
1) $\overleftrightarrow{DE}$ forms a line such that A,B,C are all on the same side of $\overleftrightarrow{DE}.$
2) A is on $\overleftrightarrow{DE}.$ Then we are done by the crossbar theorem.
3) B is on $\overleftrightarrow{DE}.$ Then we are done by the crossbar theorem.
4) C is on $\overleftrightarrow{DE}.$ Then we are done by the crossbar theorem.
5) A is on the opposite side of $\overleftrightarrow{DE}$ as B and C. Then by the plane separation theorem $\overleftrightarrow{DE}$ intersects $\overline{AB}$ and intersects $\overline{AC}.$
6) B is on the opposite side of $\overleftrightarrow{DE}$ as A and C. Then by the plane separation theorem $\overleftrightarrow{DE}$ intersects $\overline{AB}$ and intersects $\overline{BC}.$
7) C is on the opposite side of $\overleftrightarrow{DE}$ as A and B. Then by the plane separation theorem $\overleftrightarrow{DE}$ intersects $\overline{BC}$ and intersects $\overline{AC}.$
For 1), we have already determined that since D,E $\in INT(\triangle ABC)$ they are on the same side of $\overleftrightarrow{AC}$ as B. If A, B, and C were all on the same side of $\overleftrightarrow{DE},$ this would be a contradiction. So, 1) cannot be true. Therefore one of the other conditions must be true. All of the other conditions have l intersecting a side of the triangle so the theorem is proved.