Check my proof of the "Boundedness theorem" Theorem: Let $f$ be continuous on a closed interval $[a, b]$. Then f is bounded on $[a, b]$.
Proof (sketch): Suppose $f$ is unbounded. Let's define the set $N$ containing those $x$ for which $f$ is bounded on [a, x). Since $f$ is continuous there is $\delta$ such that $f$ is bounded on $[a, a + \delta)$, so $N$ is not empty. Since $N$ is bounded by $b$, we take $sup = supN$ and appealing to continuity of $f$ show that $sup$ belongs to $N$ and if $sup < b$ there is some $\delta$ such that $f$ is bounded on $[a, sup + \delta)$, so $sup + \delta$ must be in $N$ (a contradiction).
 A: The sketch looks fine for now, of course you will need to elaborate on some points, but it looks OK and it should go through. Be careful, though, as the closedness of the interval is vital and you must use it (see $1/x$ on $[-1,0)$) (also see the comment by  Martin Argerami on this topic) 2 minor points, though:
1) You do not need to work with reductio ad absurdum here. Just drop the "suppose $f$ is unbounded" at the beginning, and your proof will still hold. Basically, you first suppose $\not A$, then prove, $A$, then say "$A$ is in contradiction with $\not A$, therefore the original proposition of "$\not A$" is false, which means $A$ is true". I know this is just a minor complaint, but basically, it's simpler to just take the set $N$ and prove $N=[a,b]$, therefore proving $f$ is bounded.
2) Another, maybe simpler proof (this one DOES use contradiction): Suppose $f$ is unbounded. Then there exists a sequence $x_n$ on $[a,b]$ so that $|f(x_n)|>n$ for all $n\in \mathbb N$. Because $[a,b]$ is compact, $x_n$ has a convergent subsequence $x_{n_i}$, $i\in \mathbb N$ with limit $c$. Now, since $f$ is continuous, you know that
$$\lim_{i\rightarrow\infty} f(x_{n_i})=f(\lim_{i\rightarrow\infty} x_{n_i})=f(c),$$
however you also know that the limit does not exist. This is a contradiction, therefore $f$ cannot be unbounded.
