Let us try to rescue one more damsel-question from the unanswered questions dragon's den.
First, the roots of the quadratic equation $\,x^2-x+k=0\,$ are $\frac{1\pm\sqrt{1-4k}}{2}\Longrightarrow a:=\frac{1-\sqrt{1-4k}}{2} < h< \frac{1+\sqrt{1-4k}}{2}=:b$ Observing the geometric interpretation of the above, we have the upwards parabola $\,f(x)=x^2-x+k\,$ with two intersection points with the $\,x-$axis, both with positive abscissa, and such that $f(h)=h^2-h+k<0\,$ , since $\,f(x_0)<0 \Longleftrightarrow a .
Clearly $\,f(x_1)=f(h)<0\,$, and we also have $x_2:=x_1^2+k Thus, we see that, in general, $a
So it is enough to prove now inductively on the index of $\,\{x_n\}\,$ that $\,x_{i+1}>x_i\,\,,\,\forall i\in\mathbb N\,$ ; assuming for $\,i< n\,$ we prove it for $\,i=n$: $x_{n+1}=x_n^2+k
Thus, $\,\{x_n\}\,$ is a monotonically decreasing sequence bounded below by $\,a\,$ , so its limit exists, call it $\,\alpha\,$. Using arithmetic of limits and the recursion $\,x_{n+1}=x_n^2+k\,$ we get $\alpha\xleftarrow [\infty\leftarrow n]{}{\color{red} {x_{n+1}=x_n^2+k}}\xrightarrow [n\to\infty]{} \alpha+k\,\,\Longrightarrow \alpha^2-\alpha+k=0\Longrightarrow \alpha=a$ since the sequence decreases.