Convergence of fixed-point iteration

Step 1 (existence via the intermediate value theorem). Let $h(x)=g(x)-x$. Since $g(a)\in[a,b]$ we have $g(a)\ge a$ so $h(a)\ge0$, and similarly $g(b)\le b$ gives $h(b)\le0$. Since $h$ is continuous, the Intermediate Value Theorem gives some $r$ with $h(r)=0$, i.e. $g(r)=r$: a fixed point exists.

Step 2 (uniqueness via the mean value theorem). Suppose $r_1,r_2\in[a,b]$ are both fixed points with $r_1\ne r_2$. The Mean Value Theorem gives some $c$ between them with $g(r_1)-g(r_2) = g'(c)(r_1-r_2)$; since $g(r_1)=r_1$ and $g(r_2)=r_2$, this reads $r_1-r_2 = g'(c)(r_1-r_2)$, so $|r_1-r_2| = |g'(c)|\,|r_1-r_2| \le L\,|r_1-r_2|$. Since $L<1$ and $|r_1-r_2|>0$, this is a contradiction, so $r_1=r_2$.

Step 3 (contraction at every step). For any $x_n\in[a,b]$, apply the Mean Value Theorem to $g(x_n)-g(r)$: there is some $c_n$ between $x_n$ and $r$ with $g(x_n)-g(r) = g'(c_n)(x_n-r)$. Since $g(r)=r$ and $x_{n+1}=g(x_n)$, the left side is $x_{n+1}-r$, so $|x_{n+1}-r| = |g'(c_n)|\,|x_n-r| \le L\,|x_n-r|$.

Step 4 (iterate the contraction). Applying Step 3 repeatedly from $n=0$ gives $|x_1-r|\le L|x_0-r|$, then $|x_2-r|\le L|x_1-r|\le L^2|x_0-r|$, and inductively $|x_n - r| \le L^n |x_0 - r|$ for every $n$. Since $0\le L<1$, the right side tends to $0$ as $n\to\infty$, proving $x_n\to r$.