I am preparing a lecture on the Weierstrass theorem (probably best known as the Extreme Value Theorem in english-speaking countries), and I would propose a proof that does not use the extraction of converging subsequences. I did not explain subsequences in my calculus course, and I must choose between skipping the proof of the theorem and finding some proof which works only for functions $\mathbb{R} \to \mathbb{R}$.
I remember I once read a proof based on some bisection technique, but I can't find a reference right now.
I would be grateful for any reference to books, papers, web sites about this alternative proof.
Edit: since somebody modified my question, I will write down the precise theorem I want to prove.
Theorem. Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function. Then $f$ has at least a maximum and a minimum point.
Best Answer
Here is a proof of the Extreme Value Theorem that does not need to extract convergent subsequences. First we prove that :
Lemma: Let $f : [a,b] \rightarrow \mathbb{R}$ be a continuous function, then $f$ is bounded.
Proof: We prove it by contradiction. Suppose for example that $f$ does not have an upper bound, then $\forall n\in\mathbb{N}$, the set $\{x \in [a,b] , \, f(x) \geqslant n\}$ is not empty. Consider the following quantity: $$a_n = \inf\{x \in [a,b] , \, f(x) \geqslant n\}.$$ For all $n\in \mathbb{N}$, $a_n \in [a,b]$ exists. By the continuity of $f$, $f(a_n) \geqslant n$. And since $\{x \in [a,b] , \, f(x) \geqslant n+1\} \subset \{x \in [a,b] , \, f(x) \geqslant n\}$, we have $a_{n+1} \geqslant a_n$. Since $(a_n)_{n\in\mathbb{N}}$ is a monotonic bounded sequence, it has a limit: $$a_{\infty} = \lim_{n\to\infty} a_n $$ and $a_{\infty} \in [a,b]$. Let $M = \lceil f(a_{\infty}) \rceil$, then $\forall n \geqslant M+2$, $f(a_n) > f(a_{\infty})+1$. Hence by the continuity of $f$, $$f(a_{\infty}) = \lim_{n\to\infty}f(a_n) \geqslant f(a_{\infty})+1,$$ which yields a contradiction. Therefor $f$ must have an upper bound on $[a,b]$. For the same reason, $f$ must have an lower bound on $[a,b]$. In conclusion, $f$ is bounded on $[a,b$
With this lemma we can prove the Extreme Value Theorem.
Theorem: Let $f : [a,b] \rightarrow \mathbb{R}$ be a continuous function, then $f$ has at least a maximum and a minimum point.
Proof: We proved in the lemma that $f$ is bounded, hence, by the Dedekind-completeness of the real numbers, the least upper bound (supremum) $M$ of $f$ exists. By the definition of $M$, $$\forall n \in \mathbb{N}, \, S_n = \{x \in [a,b] , \, f(x) \geqslant M - \frac1n\} \neq \emptyset .$$ Let $s_n = \inf S_n$ be the infimum of $S_n$. We know that $a \leqslant s_n \leqslant s_{n+1} \leqslant b$ and $f(s_n) \geqslant M - \frac1n$. Since $(s_n)_{n\in\mathbb{N}}$ is a monotonic bounded sequence, the limit $s = \lim_{n\to\infty} s_n \in [a,b]$ exists. $\forall N\in\mathbb{N}$, $\forall n > N$, $f(s_n) > M - \frac1N$, so $M \geqslant f(s) = \lim_{n\to\infty} f(s_n) \geqslant M - \frac1N$. Hence $f(s) = M$ and $f$ has at least a maximum point. Similarly we can prove that $f$ has at least a minimum point. Q.E.D