So, the proof that $0<u_n<1$ is correct. It would be better to see the actual manipulations you've made, just to be sure that they are indeed correct.
Now, it's true that the sequence is decreasing (contrary to what I said in the comments, my bad), not increasing. To see this you can take $$u_{n+1}=1-\sqrt{1-u_n}\rightarrow \sqrt{1-u_n}=1-u_{n+1}\rightarrow 1-u_n=(1-u_{n+1})^2\leqslant 1-u_{n+1}$$
where in the last inequality I used $a\in[0,1]\rightarrow a^2\leqslant a$. So $1-u_n\leqslant 1-u_{n+1}$, by deleting 1 and changing sign, we reach $u_{n+1}\leqslant u_n$. In particular the limit cannot be $0$.
Now, the sequence $u_n$ is bounded (as you proved $u_n\leqslant 1$, for all $n$) and monotone (decreasing), therefore $$\exists u=\lim_{n\rightarrow \infty}u_n$$ to find the value for $u$, take the equation $$u_{n+1}=1-\sqrt{1-u_n}=f(u_n)$$ for $f(x)=1-\sqrt{1-x}$. The map $f$ is continuous, so $$u=\lim_{n\rightarrow\infty}u_n=\lim_{n\rightarrow\infty}u_{n+1}=\lim_{n\rightarrow\infty}f(u_n)=f(\lim_{n\rightarrow\infty}u_n)=f(u)$$ so $u$ is the solution to $u=f(u)=1-\sqrt{1-u}$, that is, $$\sqrt{1-u}=1-u\rightarrow 1-u=(1-u)^2\rightarrow (1-u)(1-(1-u))=0\rightarrow u(1-u)=0$$So, we have either $u=1$ (which cannot be, since the sequence is decreasing), or $u=0$, which is the correct answer. This whole process is one of the methods to find the fixed point to a continuous function $f$: You start with a value $v_0$ and the define $$v_{n+1}=f(v_n)$$
If the sequence does have a limit (as in this case with $f(x)=1-\sqrt{1-x}$), then the limit is a fixed point for $f$.
