I have moderate understanding of the lemma's use in prototypical examples like $0^n1^n$ and $WW$ (for any string $W$).
I have some confusion about the lemma's application to regular languages that don't appear to pumpable, though perhaps I'm mistaken somewhere. Suppose I define a language $L$ that can be expressed as the regular expression $1$. This language is clearly regular, because a (trivial) DFA can be constructed to accept it. The pumping lemma requires I define 3 strings, $x$, $y$, $z$, such that $L = xyz$. $z$ can be $ε$, but $y$ cannot be. There's no $xy$ I can come up with that has a $y$ that can be pumped whilst still being in my language.
This language appears to not be pumpable, and thus not regular, but I know it's regular for sure. Where's my mistake?
The pumping lemma is vacuously true for finite languages, which are all regular. If $n$ is the greatest length of a string in a language $L$, then take the pumping length to be $n+1$: trivially, if $w\in L$ and $|w|\ge p$, then the conclusion of the pumping lemma holds (as does $0=0$, and $0=1$).
The language $\{1\}$ is pumpable: all strings in the language of length $\ge 2$ can be pumped.
The pumping lemma says something about infinite regular languages (their DFAs contain loops), and can be used to prove that various (necessarily infinite) languages are not regular.