This is one of the last questions on a test and I couldn’t get the correct answer. The writing in red is my teacher’s, so my question is how would you prove it with the information my teacher has pointed out and is there any way that my original method could have worked?
On
Your teacher's comment was right, the step is assumed. However, for your method to work, it is possible, but still need to use the condition where $O, X, Z, Q$ are concyclic. In that case, the solution provided by user1430 is much better and straightforward. The following is the proof if auxiliary line $OZ$ is used.
However, on top of $OZ$, you will also need to connect $XQ$. Since $XY$ is a diameter, $∠XQY=∠XQZ=90˚$. Since $O, X, Z, Q$ are concyclic, $∠ZOY=∠XOZ=∠XQZ=90˚$. As such, $△XOZ≅△YOZ$, and hence $∠ZXY=∠ZYX$
Note that $OQ$ and $OY$ are both radius of the circle passing through $ X, Y, Q,$ and $P \implies \angle XYZ = \angle OYQ = \angle OQY $. Also, $ OXZQ$ is a cyclic quadrilateral $ \implies \angle PXO + \angle OQZ = 180^{\circ} \implies \angle PXO = \angle OQY = \angle XYZ $.
I hope it helps.