So, I'm trying to prove the following:
Let $a < b$ and let $f: [a,b] \to \mathbb{R}$ be a continuous function. If $f(a) < 0$ and $f(b) > 0$, then there exists a $c \in (a,b)$ so that $f(c) = 0$.
Proof Attempt:
Define the following set:
$$S = \{z \in [a,b]: f(z) < 0 \}$$
Since $a \in S$, $S \neq \varnothing$. Furthermore, $b$ is an upper bound for $S$. Hence, $c = \sup(S)$ exists by the completeness of $\mathbb{R}$. We will prove that $f(c) = 0$.
Let $\delta > 0$ be given. Then, I claim that there has to exist at least one element of $S$ in the interval $(c-\delta,c+\delta)$. Otherwise, all elements of $S$ lie to the left of $c-\delta$ and so, $c-\delta$ is an upper bound of $S$. That is impossible.
Suppose that $f(c) \neq 0$. If $f(c) > 0$, then the continuity of $f$ at $x = c$ tells us that:
$$\exists \delta > 0: |x-c| < \delta \implies |f(x)-f(c)| < f(c)$$
Then, we have:
$$|x-c| < \delta \implies 0 < f(x) < 2f(c)$$
which is a contradiction because we were guaranteed the existence of at least one point of $S$ in the interval $(c-\delta,c+\delta)$ and, hence, at least one point where $f(x) < 0$.
If $f(c) < 0$, then by the continuity of $f$ at $c$, we have:
$$\exists \delta >0: |x-c| < \delta \implies |f(x)-f(c)| < |f(c)|$$
$$|x-c| < \delta \implies f(c) - |f(c)| < f(x) < 0$$
which implies that there is a $z > c$ so that $z \in S$. Hence, $c$ is not an upper bound of $S$ and that is a contradiction. It follows that $f(c) = 0$, as was desired. $\Box$
Does the proof above work? If it doesn't, then why? How can I fix it?