Let $G$ be a group that has an increasing sequence of subgroups $G_i \le G_{i+1}$ satisfying the following properties.
(a) $G=\bigcup_{i\ge 1} G_i$.
(b) Each $G_i$ contains a nilpotent subgroup $N_i$ such that $[G_i:N_i] \le k$ for some constant $k$ independent of $i$.
Can we find a nilpotent subgroup $N \le G$ such that $[G:N]<\infty$?
Sorry, despite getting four upvotes, my previous answer was wrong, so I have deleted it. The result is false even when $k=1$.
As counterexample, for each $i \ge 1$, let $H_i$ be a finite nilpotent group of class $i$. For example, we could choose $H_i$ to be the dihedral group of order $2^{i+2}$.
Now let $G_i = H_1 \times H_2 \times \cdots \times H_i$ for all $i$, with the natural embedding of $G_i$ into $G_{i+1}$. Then $G := \cup_{i \ge 1} G_i = \times_{i \ge 1} H_i$ is not nilpotent, because it has subgroups that are nilpotent of class $i$ for each $i$.
But we need to prove that $G$ does not have a nilpotent subgroup of finite index. A subgroup of finite index would have to project onto $H_i$ for all but finitely many $i$, and so again it could not be nilpotent.