A={s11s|s $\epsilon$ {0}^*} so the strings 00011000 and 000001100000 are accepted of A but not 00100 or 001100000.
Demon chooses k.
I choose x = s^k, y=11, z = s^k
however i am not confident i am picking the right variables for y. am I heading down the right path?
the demon chooses u,v,w s.t. y = uvw and v does not equal $\varepsilon$
demon picks u,v,w of length j,m,n with k = j+m+n and m>0
I pick i = 2.
so xuv^2wz = s^k (and here I am stuck, do i say u is just 1 ?)
so the way I am currently learning the pumping lemma:
there exists k>=0 such that for any strings x,y,z with xyz $\epsilon$ A and |y| >= k, there exists strings u,v,w such that y = uvw, v != $\varepsilon$, and for all i>= 0, the string xuv^iwz $\epsilon$ A
The Myhill-Nerode theorem shows this immediately.
The infinitely many prefixes $0^i11$ are all pairwise distinguishable. (If $i\ne k$, then $0^i$ is a distinguishing extension: $0^i110^i \in A$ but $0^i110^k\notin A$). Therefore the language cannot be regular.
As for your attempt to use the pumping lemma, I'm not entirely sure how the formulation of the lemma you're working with goes, but you certainly can't expect any demon to split an $y$ of constant length into something that can be pumped. You must have missed some premise in the statement of the lemma.