$a_0=0$, $a_{n+1}=a_n-\frac{1}{2}(a_n ^2-a)\rightarrow \sqrt{a}$ using banach's theorem Let's have a sequence $a_0=0$, $a_{n+1}=a_n-\frac{a_n ^2-a}{2}$. This sequence converges for $0<a<1$ to $\sqrt{a}$.
One way of proving this is to prove that $T(x)=x-\frac{1}{2}(x ^2-a)$ is a contraction mapping.
$$x-\frac{x ^2-a}{2}=-\frac{1}{2}(x-1)^2+\frac{a+1}{2}$$ so for $0< a<1$ T maps $[0,\frac{a+1}{2}]\subset [1-\sqrt{1+a^2},1+\sqrt{1+a^2}]$ into $[0,\frac{a+1}{2}]$
If we define metrics on $[0,\frac{a+1}{2}]$ as $|x-y|$, then, since $\frac{\mathrm{d}}{\mathrm{d}x}(x-\frac{x ^2-a}{2})=1-x$, using mean value theorem 
$$|Tx-Ty|=|x-\frac{x ^2-a}{2}-(y-\frac{y ^2-a}{2})|=|1-\xi||x-y|$$ where $\xi=\xi(x,y) $ is between x and y.
 Since $x,y\in [0,\frac{1+a}{2}]$, $0<a<1$ $|1-\xi| $ has to be less than 1 and at least 0 so we have proven T to be a contraction mapping. Is this the correct way?

$T$ is not a contraction on $\left[0,\frac{a+1}2\right]$.
We have
\begin{align}
|Tx - Ty| &= \left|x - \frac12(x^2 - a) - y + \frac12(y^2 -a )\right| \\
&= \left|x-y - \frac12(x^2 - y^2)\right| \\
&= \left|x-y - \frac12(x-y)(x+y)\right| \\
&= \left|(x-y)\left(1 - \frac{x+y}2\right)\right| \\
&= \left|1-\frac{x+y}2\right||x-y|\\
\end{align}
If there exists $C > 0$ such that $|Tx - Ty | \le C|x-y|, \forall x,y \in \left[0,\frac{a+1}2\right]$ then in particular for $x =  0$ and $y_n = \frac1n$ we have
$$ \left(1-\frac1{2n}\right)\underbrace{|x-y_n|}_{=\frac1n} = \left|1-\frac{x+y_n}2\right||x-y_n|= |Tx -Ty_n| \le C\underbrace{|x - y_n|}_{=\frac1n}  \implies 1-\frac1{2n} \le C$$
Letting $n\to\infty$ we get $1 \le C$ so $T$ cannot be a contraction.
