Before I read the proof of Bolzano's theorem from my Calculus book, I've tried to prove it myself. I will use the following lemma and the least upper bound axiom.
[Lemma: Sign-preserving property of continuous functions.] Let $f$ be continuous at $c$ and suppose that $f(c) \neq 0$. Then there is an interval $(c – \delta , c + \delta )$ about $c$ in which $f$ has the same sign as $f(c)$.
[Bolzano's theorem] Let $f$ be continuous at each point of a closed interval $[a, b]$ and assume that $f(a)$ and $f(b)$ have opposite signs. Then there is at least one $c$ in the open interval $(a, b)$ such that $f(c) = 0$.
Proof:
(I assume that the sign of $f(a)$ is negative, and consequently $f(b)$ has a positive sign.)
1) Let $\mathbf{N}$ be the set of all points of $[a,b]$ where $f \le 0$. Then $a \in \mathbf{N}$ and $b \notin \mathbf{N}$. Since $b$ is the maximum element of the interval, $b$ is an upper bound for $\mathbf{N}$. Hence $\mathbf{N}$ has a supremum. And the supremum is strictly less than $b$ and strictly greater that $a$ (we can use the lemma to show this).
2) Let $s = sup\mathbf{N}$ and we know that $a < s < b$. Then since $f$ is continuous for every point in $[a,b]$, $f$ must be defined for every such point, and $f(s)$ has a value.
3) And I say that $f(s) = 0$. For if not, either $f(s) < 0$ or $f(s) > 0$.
4) If $f(s) > 0$ then according to the lemma there is a positive number $\delta$ such that for any value in $(s – \delta, s]$, $f > 0$. ($s – \delta$ must be greater than $a$ since $f(a) < 0$). Then none of these values from the interval is in $\mathbf{N}$. Hence for any $x \in \mathbf{N}, x \le s – \delta$. We have found an upper bound which is less than the supremum. It is a contradiction.
5) In a similar way we show that if $f(s) < 0$ than there must be numbers in $[a,b]$ bigger than $s$ for which $f < 0$, so they must be in $\mathbf{N}$.
Best Answer
(Some people use $\mathbf{N}$ for natural numbers instead of the $\mathbb{N}$ symbol mentioned in the comments, too. $N$ is better, and something like $A$ would be even more better because $N$ often stands for a number $\in \mathbb{N}$, not a set of numbers, but that's a question of taste. Also, $\sup A$ (with \sup) instead of $sup A$.)
I think the proof is correct. For what it's worth, instead of assuming $f(a) <0 $ and $f(b) >0$, with a slight modification it suffices to assume $f(a)$ and $f(b)$ have opposite signs:
Let $a' \in \{a, b\}$ such that $f(a') = \min\{f(a), f(b)\}$, and likewise $b'$ such that $f(b') = \max\{f(a), f(b)\}$.
Now clearly $f(a') < 0$ and $f(b') > 0$, and one can proceed as you've done, but working with $a'$ instead of $a$ and so on.