Let $S$ be the set of integers that represent the number of intersections of some four distinct lines in the plane. List the elements of $S$ in ascending order.
1. **Maximum Number of Intersections:**
   - Four distinct lines in the plane can intersect in at most $\binom{4}{2} = 6$ points. This is because each pair of lines can intersect at most once, and there are $\binom{4}{2}$ ways to choose 2 lines out of 4.

2. **Constructing Specific Numbers of Intersections:**
   - **0 Intersections:** If all four lines are parallel, there will be no intersections.
   - **1 Intersection:** If all four lines intersect at a single point, there will be exactly one intersection point.
   - **3 Intersections:** If three lines are parallel and the fourth line intersects each of these three lines, there will be exactly three intersection points.
   - **4 Intersections:** If the lines can be split into two pairs of parallel lines, and lines from different pairs intersect, there will be exactly four intersection points.
   - **5 Intersections:** If two lines are parallel, and the other two lines intersect each other and are not parallel to the first two lines, there will be exactly five intersection points.
   - **6 Intersections:** If no two lines are parallel and no three lines intersect at a common point, there will be exactly six intersection points.

3. **Proving 2 Intersections is Impossible:**
   - Assume for contradiction that there exists a set of four distinct lines $\ell_1, \ell_2, \ell_3, \ell_4$ that intersect at exactly two points $X_1$ and $X_2$.
   - Without loss of generality, let $\ell_1$ pass through $X_1$ and $\ell_2, \ell_3$ pass through $X_2$.
   - Since neither $\ell_2$ nor $\ell_3$ can intersect $\ell_1$ (as this would create more than 2 intersection points), both must be parallel to $\ell_1$. However, this would imply that $\ell_2$ and $\ell_3$ are parallel to each other, so they can't both pass through $X_2$, leading to a contradiction.

Hence, the possible number of intersection points for four distinct lines in the plane are $0, 1, 3, 4, 5, 6$.

The final answer is $\boxed{0, 1, 3, 4, 5, 6}$.