Theorem. Let \(f:\left[a,b\right]\rightarrow\mathbb{R}\) be a continuous function from \(\left[a,b\right]\) to \(\mathbb{R}\).
Then \(f\) is bounded on that interval.
Proof. We will first prove that \(f\) is bounded from above.
Let us assume for the sake of contradiction that \(f\) is not bounded from above. Then \[\forall M\in\mathbb{R}\ \exists x\in\left[a,b\right]\ s.t.\ f(x)>M\] and in particular, since \(\mathbb{N}\subseteq\mathbb{R}\),
\[\forall n\in\mathbb{N}\ \exists x_{n}\in\left[a,b\right]\ s.t.\ f(x_{n})>n\]
Let us look at the sequence \(\left(x_{n}\right)_{n=1}^{\infty}\). Note that \(\forall n\in\mathbb{N},a\le x_{n}\le b\), i.e. this is a bounded sequence.
Thus, the Bolzano-Weierstrass theorem states it has a convergent subsequence, \(\left(x_{n_{k}}\right)_{k=1}^{\infty}\).
Since limits of sequences weakly preserve inequalities maintained by all elements of the sequence, \[a\le\underset{k\rightarrow\infty}{\lim}x_{n_{k}}\le b\] This gives \(\underset{k\rightarrow\infty}{\lim}x_{n_{k}}=x_{0}\in\left[a,b\right]\subseteq\mathbb{R}\).
Therefore, by Heine’s characterization of continuous functions,
\[\underset{k\rightarrow\infty}{\lim}f\left(x_{n_{k}}\right)=f(x_{0})\in\mathbb{R}\]
However, notice that \[\forall k\in\mathbb{N},\ f(x_{n_{k}})>n_{k}\ge k\]
hence \[\underset{k\rightarrow\infty}{\lim}f\left(x_{n_{k}}\right)=\infty\]
This is a contradiction. This means that \(f\) is bounded from above.
To show that \(f\) is bounded from below, we can apply the same argument to \(-f\) since \(-f\) is also continuous as a scalar multiple of a continuous function.
Thus, we get the same contradiction again, hence \(f\) is bounded. ∎