$f_1(x) = \sqrt x$ and $f_{n+1}(x) = \sqrt {x + f_n(x)}$ For $x ≥ 0$, define $f_1(x) = \sqrt x$ and $f_{n+1}(x) = \sqrt {x + f_n(x)}$. 
Prove that
$(a)$ $f_n(0) = 0$ and $0 < f_n(x) < f_{n+1}(x) < 1 + x$ for all $n$ and when $x > 0$.
$(b)$ $f_n $ converges uniformly on any closed interval $[a, b]$ such that $0 < a < b$, but not on $[0,1]$.
I have proved part $(a)$ but unable to do part $(b)$ .
I think that the sequence $f_n$ converges to $1+b$  on any closed interval $[a, b]$.

Hint
Let $g(x,t)=\sqrt{x+t}$ for $x\geq0$ and $t\geq 0$.
let  $ x>0$ . By derivative,
$t \;\mapsto g(x,t) $ is increasing at $[0,+\infty)$,
and
$f_2(x)=\sqrt{x+\sqrt{x}}>f_1(x)$
thus
$(f_n(x))_n$ is  increasing.
on the other, the positive fixed point
of $t\mapsto g(x,t)$ is
$$F(x)=\frac{1+\sqrt{1+4x}}{2}>f_1(x)$$
So, by induction
$$(\forall n>0)  \;\;  0<f_n(x)\leq F(x)$$
$(f_n(x))_n$  is increasing bounded 
$\implies$ it converges to the fixed point.
finally,
$\forall x>0 $
$\lim_{n\to+\infty}f_n(x)=\frac{1+\sqrt{1+4x}}{2}=F(x)$.
and $F(0)=0$.
Observe that $F$ is discontinuous at $0$ and since all the $f_n$ are continuous at $ [0,1]$.(By induction),
$(f_n)_n$ doesn't converge uniformly at $[0,1]$.
Let us look at the convergence at $[a,b]$.
$(\forall x\in [a,b] )\;\; (\forall n\geq 1)$
$$|f_{n+1}(x)-F(x)|=$$
$$|g(f_n(x))-g(F(x))|$$
$$=\frac{|f_n(x)-F(x)|}{\sqrt{x+f_n(x)}+\sqrt{x+F(x)}}$$
$$\leq \frac{1}{2\sqrt{a}}|f_n(x)-F(x)|$$
$$\leq \frac{  1} { (2\sqrt{a })^n}|f_1(x)-F(x)|$$
$$\leq \frac{1}{(2\sqrt{a})^n}M$$
where
$$M=\sup_{x\in[a,b]}f_1(x)+\sup_{x\in[a,b]}F(x).$$
$\implies$
$$\sup_{x\in[a,b]}|f_n(x)-F(x)|\leq\frac{M}{(2\sqrt{a})^n}.$$
As $a>1$, The convergence is uniform at$ [a,b]$.
