The theorem statement is "if $f$ is continuous on $[a,b]$, $f$ is bounded on $[a,b]$". This is proven in the textbook Calculus by the author Apostol by the "method of successive bisection", which I'm sure many are familiar with. The proof is done by contradiction.
Here is my concern with this proof: we take the supremum of the leftmost endpoints of the subintervals we generate. But this assumes we can generate an infinite number of subintervals. We generated these subintervals by answering a question: "in which half is $f$ unbounded?". This is a question a person has to answer, right? So why can we do it an infinite number of times?
If let's say we stopped after a finite number of bisections $n$, we have the subintervals: $[a_n, b_n] \subset … \subset [a_1, b_1] \subset [a,b]$. Then sup$\{a, a_1, …, a_n\} = a_n$. $f$ has to be continuous at $a_n$, so we have a neighborhood $N(a_n;\delta)$ such that if $x\in N(a_n;\delta)$, then $|f(a_n) – f(x)| < \epsilon$. But we have no guarantee that $b_n – a_n < \delta$. So $f$ can be continuous at $a_n$ and still unbounded in $[a_n, b_n]$ (this doesn't result in a contradiction, which was the direction of the proof given). We can't take more bisections because the supremum, where we applied continuity, is not necessarily going to be in the smaller bisections.
Best Answer
No, you don't have to "answer" the question of in which subinterval the function is unbounded! If a function is unbounded in $A\cup B$ then it is unbounded in $A$ or unbounded in $B$ (or both). So you let $C=A$ or $C=B$, in such a way that $f$ is unbounded in $C$. Which is it, $A$ or $B$? We don't know and we don't care.
I don't follow your more detailed objection. In any case, the continuity of $f$ has nothing to do with that part of the proof; whether $f$ is continuous or not if it's unbounded on $[0,1]$ then it's unbounded on $[0,1/2]$ or unbounded on $[1/2,1]$. Continuity only comes in later, after all the bisection is done.