If $f : [a, b] \rightarrow R$ is continuous and $f(a) < 0 < f(b)$, then there is $a < c < b$ with $f(c) = 0$.
Proof.
We start to build a sequence of intervals $I_0 \supset I_1 \supset I_2 \supset \ldots$ such that $I_n = [a_n, b_n]$, and $f(a_n) < 0 < f(b_n)$ and $b_n − a_n = \frac{b−a}{2^n} .$
Let $a_0 = a, b_0 = b$ and $I_0 = [a_0, b_0]$. Then $f(a_0) < 0 < b_0$ and $b − a = \frac{(b − a)}{2^0}$. Suppose we are given $I_n = [a_n, b_n]$ with $f(a_n) < 0 < f(b_n)$ and $b_n − a_n = \frac{b − a}{2^n}$.
Let $d =\frac{b_n+a_n}{2}$. If $f(d) = 0$, then we have found $a < d < b$ with $f(d) = 0$ and are done. If $f(d) > 0$, let $a_{n+1} = a_n$ and $b_{n+1} = d$. If $f(d) < 0$, let $a_{n+1} = d$ and $b_{n+1} = b_n$. Following this, let $I_{n+1} = [b_{n+1}, a_{n+1}]$. Then $I_{n+1} \subset I_n$ and $f(a_{n+1}) < 0 < f(b_{n+1})$ and $b_{n+1} − a_{n+1} = \frac{b−a}{2^{n+1}}$.
At this point, you have to use the nested interval theorem here but I don't know how.
The intersection $\bigcap_n[a_n,b_n]$ is non empty by the nested interval theorem let $c$ be one of its element, $\lim_{n \to \infty}a_n=c=\lim_{n \to \infty}b_n$. To see this, remark that $|a_n-c|\leq (b-a)/2^n$ and $|b_n-c|\leq (b-a)/2^n$ since $c\in [a_n,b_n]$ and $b_n-a_n=(b-a)/2^n$.
This implies that $\lim_{n \to \infty}f(a_n)=f(c)$, since $f(a_n)<0$, we deduce that $f(c)\leq 0$. A similar argument shows the fact $f(c)=\lim_{n \to \infty}f(b_n)>0$ implies that $f(c)\geq 0$. This implies that $0\leq f(c)\leq 0$ thus $f(c)=0$.