Boundedness Theorem. Let $a$ and $b$ be real numbers with $a<b$, and let $f$ be a continuous, real valued function on $[a,b]$. Then $f$ is bounded above and below on $[a,b]$.

Proof. Suppose not. Then for all natural numbers $n$ we can find some $x_{n}\in[a,b]$ such that $|f(x_{n})|>n$. The sequence $(x_{n})$ is bounded, so by the Bolzano-Weierstrass theorem it has a convergent sub sequence, say $(x_{{n_{i}}})$. As $[a,b]$ is closed $(x_{{n_{i}}})$ converges to a value in $[a,b]$. By the continuity of $f$ we should have that $f(x_{{n_{i}}})$ converges, but by construction it diverges. This contradiction finishes the proof.