Clarification of the statement of the Pumping Lemma In class we were told that Pumping Lemma states:
"Let A be a regular language over $\Sigma$. Then there exists k such that for any words $x,y,z\in\Sigma^{*}$, such that $w=xyz\in A$ and $\lvert y\rvert\ge k$, there exists words $u,v,w$ such that $y=uvw$, $v\neq\epsilon$, and for all $i\ge0$ the word $xuv^{i}wz\in A$."
I was informed by the professor that one cannot make choices on what $x,y,z$ are in the above Theorem when you wish to prove a language is not regular. I'm probably missing something incredibly easy but if you assume (for the purpose of contradiction) that a language is regular does the above statement not guarantee for all choices $x,y,z$ such that $xyz\in A$ and $y$ is sufficiently long that the conclusion of the Pumping Lemma holds? Perhaps a quantifier is missing?
Any clarification to the statement of the Pumping Lemma would be appreciated. Thank you very much for your help.

Your understanding is correct. If you want to apply the lemma to a language you know (or assume) is regular, you can choose $x$, $y$, and $z$ any which way you fancy, as long as you make sure $y$ is long enough. (But since you get $k$ from the lemma, what you actually need in the usual scenario is to provide is a procedure that explains how you're going to choose $x$, $y$, and $z$ after you know what $k$ is).
In light of this, your professor's statement is rather peculiar. Perhaps he was referring to proving the lemma? If your task is to prove the lemma rather than apply it, the burden of proof switches around, and you now have to find a way to deal with whatever $xyz$ your adversary shows up with (after you tell him a $k$).
