for all integer a>1 there can be found at least one prime divisor

i am starting to study abstract algebra and at this question i missed some points of the proof i wish you can explain:
prove that-->for all integer a>1 there can be found at least 1 prime divisor.
proof:let S={d (element) Z: d>1 and d|a} be this set.we know that S is not empty since a|a and a>1 so a (element) S and Sis not empty set.
positive integer set is well-ordered set so there exist the minimum element of S and say it p.if we show that p is prime number then the proof is over.
from here;
assume contrary that p is not prime then p=q.r and there can be found at least one q,r (element) Z such that 1<q<p.
and q|p and p|a -->q|a and q (element) S and this contradicts witht the assumption of p be the smallest element of S since p is prime.

my question is why do we try to prove that p is smallest element of S.we are searching at least one prime divisor why that must be the smallest one,what happened if the smallest one is not prime?may there is another elemnt is prime of S.

You dont prove p is the smallest term. The fact is that since S in contained in the positive integers then there exists a least term due to the well ordering principle. So if you take this least term and do all this stuff...

...if we show that p is prime number then the proof is over. from here; assume contrary that p is not prime then p=q.r and there can be found at least one q,r (element) Z such that 1<q<p. and q|p and p|a -->q|a and q (element) S and this contradicts witht the assumption of p be the smallest element of S since p is prime...

Then you show that this least element has prime divisors, hence if you take ANY set S containing ANY elements in the positive integers then the least term has prime divisors, hence every element in the positive integers has prime divisors.

For example take the set A that is the set S without S's least term, then by applying the same argument as above the least element of A has prime divisors...
Then the set B = A without A's least term, same argument... So B's least term has prime divisors, etc...

This is how I would write out my proof.

Suppose to the contrary that there is an integer greater than 1 that is not divisible by any prime. Let be the smallest such integer. Then must be composite (if it were prime, then it would be divisible by a prime, namely itself). Thus where By minimality of there is an prime which divides It follows that divides so that is divisible by a prime after all. This contradiction means that no such integer exists; hence every integer greater than 1 is divisible by some prime.