the following attempted proof of the parallel postulate is s

the following attempted proof of the parallel postulate is similar to Proclus\' but the flaw is different; detect the flaw with the help of exercise 1. Given P not on line l, pq perpendicular to l at q, and line m perpendicular to pq at p. let n be any line through p distinct from m and pq. We must show that n meets l. Let px be a ray of n between pq and a ray of m emanting from p and let y be the foot of the perpendicular from x to pq. As x recedes endlessly from p, py increases indefinitely. Hence, y eventually reaches a position y\' on ray pq such that py\'>pq. Let x\' be the corresponding position reached by x on the line n. New x\' and y\' are on the same side of l because line x\'y\' is parallel to l. But y\' and p are on oppsite sides of l. Hence, x\' and p are on opposite sides of l, so that segment px\' meets l.

Solution

Given P not on line l, pq perpendicular to l at q, and line m perpendicular to pq at p. let n be any line through p distinct from m and pq. We must show that n meets l. Let px be a ray of n between pq and a ray of m emanting from p and let y be the foot of the perpendicular from x to pq.

Till this point it is fine. Also As x recedes endlessly from p, py increases indefinitely is true. But this increase is not bounded is not accomplished. Because a sequence may increase indefinitely yet it may be bounded by a fixed number and may not cross that number at any time. For example, 1/2, 3/4, 7/8, 9/16, 31/32, 63/64,.. is increasing but it never crosses the point 1. Thus saying that y eventually reaches a position y\' on ray pq such that py\' >pq requires justification.  

Hence the argument is not complete. It is assuming that the distance between the two lines l and m is constant and also assuming that it can be crossed eventually.