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The hint that I've been given is: for each n in the naturals, use the assumption that $n$ is not an upper bound for $f(I)$ to choose a sequence of $x_n$ (from $n=1$ to infinity) in $I$; then apply Bolzano-Weierstrass.

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Yes, exactly, do so. – Hagen von Eitzen Dec 9 '12 at 16:32
@HagenvonEitzen I have no idea how to go about it though? I don't even understand the hint. – Mathlete Dec 9 '12 at 16:34
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Suppose that $f$ is not bounded above that is \begin{equation}\forall M\in \mathbb{R} \exists x\in [a,b]:f(x)>M\end{equation} Taking $M=1,...,n,...$ we can construct inductively a sequence $(x_n)\in [a,b]$ such as that \begin{equation}\forall n\in \mathbb{N}\;f(x_n)>n\end{equation} The sequence $(x_n)$ is bounded and so by the Bolzano-Weierstrass Theorem it has a convergent subsequence $x_{k_n}\to c$ Since $\forall n\in \mathbb{N}\ a\le x_n \le b$ we have that $a\le c \le b$. By hypothesis, $f$ is sequentially continuous at $c$ and because $x_{k_n}\to c$ it follows \begin{equation}f(x_{k_n})\to f(c)\end{equation} which is a contradiction to the fact that \begin{equation}\forall n\in \mathbb{N}\;f(x_n)>n\end{equation} Thus, $f$ is bounded above.

EDIT: We have $\forall n\in \mathbb{N}\;f(x_n)>n$. This implies \begin{equation}\lim_{n\to \infty}f(x_n)=+\infty\end{equation} which is a contradiction to the fact that $f(x_{k_n})\to f(c)$

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Are you assuming that the sequence x_n is bounded above? – Mathlete Dec 9 '12 at 16:52
It is bounded above by say, $b$. – Nameless Dec 9 '12 at 16:52
Oh of course. How is the the fact that f(x_k_n) tending to f(c) a contradiction to the following fact? – Mathlete Dec 9 '12 at 16:53
Take the limit of the inequality as $n\to +\infty$ – Nameless Dec 9 '12 at 16:54
What do you mean? – Mathlete Dec 9 '12 at 16:56

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