Question regarding standard proof to: If f is continuous on [a,b] then f is bounded on [a,b]. The standard proof first builds up a sequence by saying that for each integer n there exist a number $x_n$ in [a,b] such that |$f(x_n)$| > $n$. Then, it proceeds to use the Bolzano weierstrass theorem to reach a contradiction. I understand that part and the rest of the proof but never really understood why this sequence $x_n$ Mwas bounded (since we rely on the fact it's bounded to use the BW Theorem). Can anyone give me some help here? Thank you!

I assume the proof begins with the assumption that $f$ is continuous but not bounded. If it is not bounded, then no matter what bound we try to put on it (our $n$), we can always find a point that exceeds it (our $x_n$).
It's not the $f(x_n)$ that are bounded (because, by assumption, they aren't), it's the $x_n$. And since $[a,b]$ is compact, this subsequence converges to a point in $[a,b]$. Then the contradiction follows neatly.
