Let $f: [a,b] \to\mathbb{R}$ be a continuous function such that $f(a) < 0 \leq f(b)$.
Show that the bisection method converges to a zero of $f$. That is, show that $\bigcap\limits_{i=0}^{\infty} I_{i} = \{x_0\}$ for some $x_0\in [a,b]$ and $f(x_0) = 0$.
I really don't know where to go from here, I've tried using the iterating method for the bisection method, namely:
Iterating the intervals $I_i = [a_i,b_i]$ such that $I_0 = [a,b]$;
And then for every iteration $I_{i+1} = [a_{i+1},b_{i+1}] = \begin{cases} [a_i, \frac{a_i + b_i}{2}], & \text{if } f(\frac{a_i + b_i}{2}) >0, \\ [\frac{a_i + b_i}{2}, b_i], & \text{otherwise}.\\ \end{cases}$
After doing this though I have no clue what to do next.
If anyone had advice I'd gladly take it, thanks.
You want the Nested Interval Theorem. This tells you that if you have a sequence of nested closed and bounded intervals, where each interval is a subset of the previous one (like in your case), the infinite intersection is non-emtpy, so the intersection you have defined will contain some points.
Now you have to ask yourself, in your case, what points can it contain? Suppose a point $x$ is in the infinite intersection. Either $f(x)$ is positive, it is negative, or it's zero. If if is positive, you should be able to show that eventually $x$ will be left out of one of the intervals, and therefore it cannot be on the intersection, yielding a contradiction. If it's negative, you argue in the same way. The only possibility left is that $x$ is such that $f(x)=0$. You can check that specific $x$ will be inside all of your intervals, and therefore also in the intersection.
Note you might also what to think about what happens if $f$ has multiple zeros on the interval $[a,b]$.
Edited for clarity.