Please let me know if the following proof is solid or if it contains errors. Thanks.

Show that every quadrilateral (convex or not) has a side \overline{XY} such that the other two vertices lie on the same side of the line \overleftrightarrow{XY}. (From Edwin E. Moise Elementary Geometry from and Advanced Standpoint 3rd Ed.)

Proof: Given quadrilateral WXYZ. We have segments \overline{WX} , \overline{XY} , \overline{YZ} , and \overline{ZW} that intersect at the corners of the quadrilateral.

Now by Incedence Axiom 1, segment \overline{ZW} has points Z and W which are contained in one and only one line, \overleftrightarrow{ZW} . By incidence axiom 3 and the definition of a quadrilateral, this line is entirely contained in the plane containing the quadrilateral.

By the plane separation theorem, this line divides the plane containing the quadrilateral into two half-planes.

Now consider the segments \overline{WX} and \overline{YZ} . Since W is on the line, X must lie in one half-plane. Similarly, since Z is on the line, Y must lie in one half-plane. So X and Y must lie in either the same half-plane or opposite half-planes determined by the line \overleftrightarrow{ZW} .

If X and Y lie in opposite half-planes, then the side of the quadrilateral, \overline{XY} , would have to intersect the line \overleftrightarrow{ZW} by the plane separation theorem. If \overline{XY} intersects \overline{ZW} , this is a contradiction which proves that X amd Y lie in the same half-plane which proves that W and Z lie on the same side of the line \overleftrightarrow{XY} .

If \overline{XY} intersects the line \overleftrightarrow{ZW} outside the segment \overline{ZW} , then Z and W lie in the same half-plane determined by the line \overleftrightarrow{XY} .