When proving a set is not regular is it enough to prove a subset of it regular?
I think you mixed up a few things here. Say, L is indeed regular. Then the pumping lemma guarantees us a pumping length. 

If we want to show that $L$ is not regular, we can go on by assuming its regular and for every pumping length we find a word that cant be pumped.

This means, we need to search for a word (for example $ab^pab^pa$ as you have suggested) and show that no matter how you divide this word into the 3 pieces $x,y,z$ ($w=xyz$) of the pumping lemma, you can find an integer $n$ with $xy^nz\notin L$. This would contradict the pumping lemma and therefore the regularity of $L$.

As a side note, showing that a subset of $L$ is not regular doesnt imply that also $L$ is not regular (for example, $\Sigma^*$ is regular, but there are non-regular languages which are obviously subsets of it).