Plsea check my proof for the fact mentioned above.
We use two lemmas which are proved befero wiohtut proof (Let $I = [0,1]$ and $f: I \rightarrow \mathbb{R}$ be a continous function)
- Let $a_1, a_2, \cdots, $ are a nondecreasing sequence of reals inside $I$, then it has a limiting value $\alpha$ and morever $\alpha \in I$.
- For any set of reals $S \in I$, the $\inf S$ and $\sup S$ exists and $\inf S, \sup S \in I$
Suppose for the sake of contradiction, $f$ is not bounded above (the bounded below case is similar). Now, let $S_i = \{ x \in I | f(x) \geq 2^i \}$. By our assumption, $S_i$ is nonempty, so by (2) it has a infimum $a_i$.
Lemma: $f(a_i) \geq 2^i$
Proof: Assume the condriction, so by continuity there's a neighbourhood $Q = [a_i - \epsilon, a_i + \epsilon] \cap I$ of $a_i$ such that $f(x) < 2^i$ for all $x \in Q$. But then the infimum can't be inside $Q$, so the infimum is greater than $\min (a_i + \epsilon, 1)$, so $a_i > \min (a_i + \epsilon, 1)$, a contradiction.
Also, since $S_i \subset S_j$ for any $i < j$, the sequnece $a_1, a_2, \cdots, $ are nondecreasing and has a limiting value $\alpha \in I$ by (1).
Now,by the definition of limits, for every $\epsilon > 0$, we can find an integer $N$ such that for all $m > N$, we have $|\alpha - a_m| < \epsilon$. Also, combining it with the definition of continuity, for every $\epsilon > 0$, we can find an $N$ such that for all $m > N$, we have $|f(\alpha) - f(a_m)| < \epsilon$.
Now we're done - coupled with the lemma it tells that $ f (\alpha) \geq f(a_m) + \epsilon \geq 2^m + \epsilon$ for all sufficiently large $m$ and sufficiently small $\epsilon$, which is a clear contradiction isnce $f(\alpha)$ is finite.
This can be done by a much simpler method for the sake of contradiction of the boundness of f. Assume a sequence defining by $f(x_n)>n$ now $x_n$ sequence is bounded so there exist a subsequence of $x_n$ is converges to some point on that closed interval. Now simply it is clear the some subsequence of $f(x_n)$ converges.