>I pick a < p ,proved that do not exist a , b , c >rational numbers with 0<a=<b<c and a<p to satisfy >a^p+b^p=c^p
No, you never proved the above claim.
You only thought you did.
First you tried to show that the equation
a^p + b^p = c^p
has no solutions in integers a,b,c,p subject to the conditions 0 < a <= b < c, p > 2, a < p.
For the sake of argument, let's allow that claim.
Then you attempted to extend to positive rationals a,b,c. To do that, you scale a,b,c down, dividing each by a positive integer large enough so the new value of a is less than p. Then you claim a contradiction since you already showed that a < p is impossible. But you showed that for positive integer values of a, not for positive rational values of a, so (barring circular reasoning) you don't have your claimed contradiction.