The iteration defined by $x_{n+1} = \frac12 (x_n^2 + c)$ where $0< c < 1$ has two fixed points $r_1$ and $r_2$ where $0<r_1<1<r_2$. Prove that $x_{n+1} - \xi_1 = \frac12 (x_n + r_1) (x_n -r_1), n =0,1,2,...$ and deduce the $\lim\limits_{n \to \infty} x_n = r_1$ if $0 \leq x_0 < r_2$.
I am not getting any clue how to start. Please Help.
First, you have a typographical error: The $\xi_1$ should be $r_1$. Here are hints to get you going:
To find the fixed points of the iteration, you have to solve the equation $r=\frac12(r^2+c)$. Use the quadratic formula to obtain two roots, call them $r_1$ and $r_2$ with $r_1<r_2$. Notice that $r_1+r_2=2$.
To establish the identity $$ x_{n+1}-r_1 = \frac12(x_n+r_1)(x_n-r_1),\tag1$$ rewrite the LHS of (1) by substituting $x_{n+1}:=\frac12(x_n^2+c)$, and $r_1=\frac12(r_1^2+c)$ (why can you assert this?)
By the same reasoning you can also establish $$ x_{n+1}-r_2 = \frac12(x_n+r_2)(x_n-r_2).\tag2$$
To finish the proof, assume that $0\le x_0<r_2$. You can use (2) to prove by induction that $0\le x_n<r_2$ for all $n$, and therefore $0\le \frac12(x_n+r_1)<\frac12(r_2+r_1)$ for all $n$. But $r_2+r_1=2$, so you've just shown that the factor $\frac12(x_n+r_1)$ in identity (1) is nonnegative and strictly less than $1$. This fact will be useful later.
Now argue in three cases: (a) $0\le x_0<r_1$, (b) $x_0=r_1$, (c) $r_1<x_0<r_2$. In case (b) it is clear that $x_n$ remains constant at $r_1$. In case (a) you have enough information to show that $\{x_n\}$ is an increasing sequence and bounded above by $r_1$, hence converges [hint: induction using (1)]. In case (c) you can similarly show that $x_n$ is decreasing and bounded below by $r_1$, hence converges. Now the limit in all cases will be a fixed point of the iteration, so it's either $r_1$ or $r_2$; you have to argue why it must be $r_1$.