= Solution
Every $x_0\in[1,\infty)$ works. Indeed, $x_1\geq\sqrt a$, and whenever $x_k\geq\sqrt a$,
$$
0\leq x_{k+1}-\sqrt a
=\frac{(x_k-\sqrt a)^2}{2x_k}
\leq x_k-\sqrt a.
$$
The iterates from $k=1$ onward therefore decrease to a <limit>, and the recurrence forces that <limit> to be $\sqrt a$.
There is also an explicit error bound. Put
$$
q=\left|\frac{x_0-\sqrt a}{x_0+\sqrt a}\right|<1.
$$
A direct calculation gives
$$
\frac{x_{k+1}-\sqrt a}{x_{k+1}+\sqrt a}
=\left(\frac{x_k-\sqrt a}{x_k+\sqrt a}\right)^2.
$$
Hence, for $k\geq1$,
$$
\boxed{
|x_k-\sqrt a|
=\frac{2\sqrt a\,q^{2^k}}{1-q^{2^k}}
\leq\frac{2\sqrt a}{1-q}q^{2^k}}.
$$
This double-exponential decay is consistent with the Newton bound in part (a). For the chosen objective, the supplied factorization also gives
$$
f(x)-f(\sqrt a)
=\frac13(x-\sqrt a)^2(x+2\sqrt a),
$$
so convergence of the objective and convergence of the iterates are equivalent on $[1,\infty)$.
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