Proving a language with equal occurences of ab, and cd is not a regular language using the Pumping Lemma
Ok I will split the answer into two parts, one for each of your questions \ concerns \ confusions:

Q: Are we allowed to use the pumping lemma if we chose a specific decomposition, or do we need to show for all decompositions?
A: The pumping lemma states, $\exists p.\forall w.|w|\ge p \rightarrow \exists x,y,z.w=xyz \land\text{ they are a valid decomposition }\land\forall k\in\mathbb{N}.xy^kz\in A$.
and you are trying to show the negation of this statement. as you can see, there is an &quot;exists&quot; sign before the &quot;x,y,z&quot; and therefore in the negation there will be a &quot;for all&quot; sign.
Thus, you MUST go over ALL possible decompositions in order to use the pumping lemma in this way.

Q: This doesn't work for $p=1$ but works for every other value of $p$. is this a problem?
A: no! its absolutely fine.
Say, for example, you know how to prove for $p=5$ and you don't know how to do $p=1$. define $w=(ab)^5(cd)^5$, and you know that $|w|\ge 5 \ge 1=p$. but since $|w|\ge 5$ and you know how to solve for $p=5$, you know how to deal with $w$.
This means that its enough to prove for some value of $p$ and all other values BELOW it will be automatically proved too!

Also, you are in the right direction. This is a good choice of a word, but you will need to work a bit more to show that every decomposition cannot be always pumped.
I hope this helped you a bit :)