= 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 $|f(x_n)|\geq n$. 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 $|f(x_{n_k})|\geq n_k\to\infty$. Thus \b[$f$ is bounded].
Let $s=\sup_{x\in[a,b]}f(x)$. By the definition of the <supremum>, choose $y_n$ with $s-1/n<f(y_n)\leq s$. 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 \b[both the supremum and the infimum are attained]. This proves the <extreme value theorem> for a closed bounded interval.
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