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.
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
Note: This answers the question as it was originally asked, which unfortunately wasn’t what was intended. I’ve added a sketch of an answer to the intended question below it.
I don’t have a reference at hand, but the proof is straightforward. Suppose that $f$ is continuous on $[a_0,b_0]$ with $f(a_0)<0<f(b_0)$. Recursively construct a sequence of intervals $[a_k,b_k]$ as follows.
If $f\left(\frac{a_k+b_k}2\right)=0$, stop: you’ve found the desired point.
If $f\left(\frac{a_k+b_k}2\right)<0$, let $a_{k+1}=\dfrac{a_k+b_k}2$ and $b_{k+1}=b_k$.
If $f\left(\frac{a_k+b_k}2\right)>0$, let $a_{k+1}=a_k$ and $b_{k+1}=\dfrac{a_k+b_k}2$.
Then $\langle a_k:k\in\Bbb N\rangle$ is non-decreasing, $\langle b_k:k\in\Bbb N\rangle$ is non-increasing, $a_k<b_m$ for all $k,m\in\Bbb N$, and $b_k-a_k=(b-a)2^{-k}$ for all $k\in\Bbb N$, so $\sup_ka_k=\inf_kb_k$, and by continuity the function must be $0$ at this point.
Obviously you’d want to fill in a lot of detail and show how it generalizes to the full intermediate value theorem, but that’s it in a nutshell.
To use bisection to prove the extreme value theorem, let $f$ be a continuous real-valued function on $[a,b]$. First use bisection to show that $f$ is bounded above: if not, at stage $k$ choose the left half if it contains a point $x_k$ such that $f(x_k)\ge k$, and otherwise choose the right half and a point $x_k$ in it such that $f(x_k)\ge k$. Then $a_k\le x_k\le b_k$ for all $k\in\Bbb N$, so $\langle x_k:k\in\Bbb N\rangle\to\sup_ka_k$, and $f$ cannot be defined at $\sup_ka_k$.
Now let $y=\sup_{x\in[a,b]}f(x)$, and repeat the bisection process, but this time at stage $k$ choose the left half if it contains a point $x_k$ such that $f(x_k)>y-2^{-k}$; if not, the right half contains such a point $x_k$, and we choose the right half. Then $\langle x_k:k\in\Bbb N\rangle\to\sup_ka_k$, and $f(\sup_ka_k)=y$.
Here’s an outline of a “bisection” proof.
Suppose $\sup_{x \in [a,b]} f = \alpha$. Define a sequence of intervals $I_k = [a_k,b_k]$ recursively as follows: $I_0 = [a,b]$ and for $k > 0$
$
I_k =
\begin{cases} [a_{k-1}, \frac{a_{k-1} + b_{k-1}}{2}] \quad \text{if }
\sup_{x \in [a_{k-1},\frac{a_{k-1} + b_{k-1}}{2}]} = \alpha \\
[\frac{a_{k-1} + b_{k-1}}{2},b_{k-1}] \quad \text{otherwise}.
\end{cases}
$
Then there exists $x \in \cap_{k} I_k$ (though this requires some care to show, and is a theorem by itself), and you can show $f(x) = \alpha$.