 Solution

We use the Bolzano-Weierstrass theorem: every bounded real sequence has a convergent subsequence. If $f$ were unbounded on $[a,b]$, there would be $x_n\in [a,b]$ with $|f(x_n)|\ge n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_k}\to c$, and the closed interval contains $c$. Since $f$ is a continuous function, $f(x_{n_k})\to f(c)$, contradicting $|f(x_{n_k})|\ge n_k\to \infty$. Thus $f$ is bounded.

Let $s={\sup }_{x\in [a,b]}f(x)$. By the definition of the supremum, choose $y_n$ with $s-1/n<f(y_n)\le s$. Again the Bolzano-Weierstrass theorem gives $y_{n_k}\to d\in [a,b]$. Continuity yields $f(d)=s$. Applying the same argument to $-f$ gives a point where $f$ attains its infimum. Therefore both the supremum and the infimum are attained. This proves the extreme value theorem for a closed bounded interval.