smc

Note that terms of the sequence $(u_k)$ lie in the interval $\left(0,\frac{1}{2}\right),$ strictly increasing. Since the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$ The given equation becomes $$L=2L-2L^2,$$ from which $L=\frac{1}{2}.$ The given inequality becomes $$\frac{1}{2}-\frac{1}{2^{1000}}\le u_k\le \frac{1}{2}+\frac{1}{2^{1000}},$$ and we only need to consider $\frac{1}{2}-\frac{1}{2^{1000}}\le u_k.$ We have \begin{alignat}{8} u_0 &= \phantom{1}\frac14 &&= \frac{2^1-1}{2^2}, \\ u_1 &= \phantom{1}\frac38 &&= \frac{2^2-1}{2^3}, \\ u_2 &= \ \frac{15}{32} &&= \frac{2^4-1}{2^5}, \\ u_3 &= \frac{255}{512} &&= \frac{2^8-1}{2^9}, \\ & \phantom{1111} \vdots \end{alignat} By induction, it can be proven that $$u_k=\frac{2^{2^k}-1}{2^{2^k+1}}=\frac{1}{2}-\frac{1}{2^{2^k+1}}.$$ We substitute this into the inequality, then solve for $k:$ $$\begin{array}{rl}\frac{1}{2}-\frac{1}{2^{1000}}& \le \frac{1}{2}-\frac{1}{2^{2^k+1}}\\ -\frac{1}{2^{1000}}& \le -\frac{1}{2^{2^k+1}}\\ 2^{1000}& \le 2^{2^k+1}\\ 1000& \le 2^k+1.\end{array}$$ Therefore, the least such value of $k$ is $\boxed{10}.$