Bisection convergence and error bound

Step 1 (the invariant). Set $a_0=a$, $b_0=b$. At step $n$, we maintain the invariant that $f(a_n)f(b_n)<0$, i.e. the root is trapped somewhere in $[a_n,b_n]$. This holds initially by hypothesis. Given $[a_n,b_n]$, compute the midpoint $x_n=\dfrac{a_n+b_n}{2}$ and evaluate $f(x_n)$: if it has the opposite sign to $f(a_n)$ set $a_{n+1}=a_n,\ b_{n+1}=x_n$, otherwise set $a_{n+1}=x_n,\ b_{n+1}=b_n$ (if $f(x_n)=0$ exactly, $x_n$ is the root and the process stops). Either way $f(a_{n+1})f(b_{n+1})<0$ again, so the invariant is preserved by induction.

Step 2 (shrinking width). By construction each new interval is exactly half the previous one, so $b_n-a_n = \dfrac{b-a}{2^n}$ for every $n \ge 0$. Since the root $r$ lies in $[a_n,b_n]$ at every step (by Step 1), and the midpoint $x_n$ also lies in $[a_n,b_n]$, both are within the same interval of width $\dfrac{b-a}{2^n}$ of each other, so $|x_n - r| \le \dfrac{b-a}{2^n}$ already holds; a slightly sharper count (measuring from the midpoint to either endpoint) gives the stated bound $|x_n - r| \le \dfrac{b-a}{2^{n+1}}$.

Step 3 (convergence). Since $\dfrac{b-a}{2^{n+1}} \to 0$ as $n\to\infty$, the squeeze from Step 2 forces $x_n \to r$. No continuity of $f$ beyond the initial sign condition is even needed for this part — only for guaranteeing a root existed inside $[a,b]$ to begin with, which is the Intermediate Value Theorem applied at step 0.