Solutions

For contradiction, suppose that no midpoint $y$ of any interval $I_n$ is such that $f(y)=0$. Without loss of generality, consider

$$y\in\bigcap_{i=1}^{\infty} I_i$$

with $f(y)<0$. There exists $\delta>0$ such that $f(x)<0$ whenever $|x-y|<\delta$. Notice that the length of $I_n$ is given by

$$\frac{b-a}{2^n}.$$

We can find $n\in\mathbb{N}$ such that $y\in I_n$ and

$$\frac{b-a}{2^n}<\delta.$$

This would mean $I_n$ only contains numbers $z$ such that $f(z)<0$. But this contradicts the definition of these intervals.