Of course, whenever the Pumping Lemma for Regular Languages (or CFLs) is applied to a language $L$, only a single string $w \in L$ and $|w| \ge p$ needs to be picked (to eventually show that $w$, when pumped, leaves the language).Is there a non-regular (or non-context-free) language $L$ for which every string $w \in L$ having $|w| \ge p$ can be used to derive a contradiction?
Is there a language for which any string (that works) can be shown to not be pumped using the Pumping Lemma?
formal languages;pumping lemma
I'd say the language $L = $ { $0^n1^n | n > 0$ } fits the bill for the Pumping Lemma for Regular Languages.  Any substring to be pumped would consist either of all the same symbol, in which case pumping up produces a string with an unequal number of $0$'s and $1$'s, or it contains instances of both symbols, in which case pumping up produces a string with a $1$ followed by a $0$; in neither case is the pumped-up string in $L$.