In a calculus course we prove a lot of theorems as fundamental consequences of the least upper bound property of R. In some cases, such as the Intermediate Value Theorem and the Bolzao-Weierstrass theorem, they turn out to be equivalent.
My question is if the same applies to the fact that every function that is continuous on a closed interval is bounded on that interval.
That is: if F is an ordered field satisfying the above fact, is F a complete ordered field?
From the Extreme Value Theorem it's easy to prove Rolle's Theorem and then the Mean Value Theorem, and this answer shows that we then have the least upper bound property. But I'm asking about the more primitive fact, where we only assume boundedness of continuous functions on closed intervals (and not necessarily that they have a maximum).
Below I'll write a (pretty long) proof that if we assume both the property above, together with the Archimedean property, then we can prove what I want. I wonder if the proof has any mistakes, and either way, if there is an easier proof, maybe one not using the Archimedean property.
TLDR Summary: Given an ordered field such that every function which is continuous on a closed interval is bounded there, can we conclude that F is complete (without assuming the Archimedean property)?
My proof (EDITED)
Suppose $F$ is an ordered field satisfying the Archimedean property, such that every function which is continuous on a closed interval is bounded on that interval. Assume in way of contradiction that $F$ is not a complete ordered field. It's possible to conclude that there is a monotone sequence which is bounded but not convergent. We can assume WLOG that there is a strictly increasing sequence $(a_n)_{n=1}^\infty$ which is bounded above but has no limit.
Let $A = \{ a_n|n\in \mathbb{N} \}$ and let $B$ be the set of all upper bounds of $A$. Note that $B$ is not empty, and $A$ has no least upper bound (otherwise $(a_n)$ would converge to $\sup A$).
Define a function $f$ on $[a_1, \infty)$ as follows.
If $x \in B$ (and in particular $x \geq a_1$) then $f(x) = 0$.
If $x \notin B$ but $x \geq a_1$ then we can find a unique $n$ such that $ a_n \leq x < a_{n+1} $, and define $$ f(x) = \frac{1}{a_{n+1} - a_n} (x - a_n) + n.$$
First, if $x \in B$ then, since $x$ is not the least upper bound of $A$, there is some $y<x$ in $B$, so $f$ is constant on $(y, \infty)$ and in particular continuous at $x$.
Clearly $f$ is continuous on $(a_n, a_{n+1})$ for all $n$, and a straight forward check of one-sided limits shows that $f$ is continuous on $[a_1, \infty)$ (it was done with more detail in the old proof).
Let $M$ be some element of $B$. Then $f$ is continuous on $[a_1, M]$, but, it's not bounded on that interval by the Archimedean property, since $f(a_n) = n$ for each $n$.
My proof (OLD)
A little after writing my original question I thought of a better idea. For the sake of documentation, here is my first proof:
Suppose $F$ is an ordered field satisfying the Archimedean property, such that every function which is continuous on a closed interval is bounded on that interval, and that $A$ is a nonempty subset of $F$ which is bounded above.
Let $B_+$ be the set of all upper bounds of $A$, and $B_-$ be the set of all elements of F which are not upper bounds for $A$. Since $A$ is nonempty, we can find some $a_0$ in $B_-$, and since $A$ is bounded above, we can find some $b_0$ in $B_+$.
We can now recursively construct a nondecreasing sequence $(a_n)$ and a nonincreasing sequence $(b_n)$ such that, for all $n$, $a_n \in B_-$, $b_n \in B_+$ and $b_n - a_n = \frac{b_0-a_0}{2^n}$ (the construction can be found in any proof that Cauchy completeness with the Archimedean property imply the least upper bound property). By the Archimedean property, $\lim_{n \rightarrow \infty} (b_n-a_n) = 0$.
Assume in way of contradiction that $A$ has no least upper bound. It's not hard to show that if $(b_n)$ converges, then so does $(a_n)$, and to the same limit, and that this limit is a least upper bound for $A$. Thus $(b_n)$ does not converge, and, in particular, it's not eventually constant. Therefore we can find an increasing sequence $(n_k)$ of natural numbers such that $(b_{n_k})$ is increasing, and $n_0 = 0$.
Suppose, in way of contradiction, $x$ is an upper bound for $A$ such that $x \leq b_n$ for all n. Then $0 \leq x - a_n \leq b_n - a_n = \frac{b_0-a_0}{2^n}$, so, again by the Archimedean property, $(a_n)$ converges to $x$. It follows that $(b_n)$ also converges, a contradiction. Therefore if $x \in B_+$ then there is some $n$ such that $b_n < x$.
We now define a function $f$ on the interval $(-\infty, b_0]$. For convenience denote $$y_k=\frac{1}{b_{n_k} - a_{n_k}}=\frac{2^{n_k}}{b_0-a_0}$$ and also $x_k=b_{n_k}$.
If $x \in (-\infty, b_0] \cap B_+$ then there is a unique $k$ such that $x_k< x \leq x_{k-1}$; let $$f(x) = \frac{y_k - y_{k-1}}{x_k - x_{k-1}} (x - x_{k-1}) + y_{k-1}.$$ If $x \in B_-$ let $f(x) = 0$.
First, $f$ is clearly not bounded on $[a_0, b_0]$: for all k, $x_{k-1}$ is in that interval and $$f(x_{k-1}) = y_{k-1} =\frac{2^{n_{k-1}}}{b_0-a_0}.$$ Once again by the Archimedean property, this can be made arbitrarily large by picking $k$ sufficiently large.
We just need to show $f$ is continuous on that interval.
First, if $x \in B_-$ then there is some $a$ in $A$ such that $x < a$, so $(x-1,a) \subseteq B_-$, so $f$ is constant on the interval $(x-1,a)$, which contains $x$, so $f$ is continuous at $x$.
Clearly $f$ is continuous on $(x_k, x_{k-1}]$ for each k, and it just remains to show that $f$ is continuous at $x_k$ for each $k$. Well, $x_{k+1} < x_k \leq x_k$, so $$f(x_k) = \frac{y_{k+1} - y_k}{x_{k+1} - x_k} (x_k - x_k) + y_k = y_k,$$ and $$\lim_{x \rightarrow x_k^-} f(x) = \lim_{x \rightarrow x_k^-} \left(\frac{y_{k+1} - y_k}{x_{k+1} - x_k} (x - x_k) + y_k \right) = y_k$$ while $$\lim_{x \rightarrow x_k^+} f(x) = \lim_{x \rightarrow x_k^+} \left(\frac{y_k - y_{k-1}}{x_k - x_{k-1}} (x - x_{k-1}) + y_{k-1}\right) = y_k$$ and therefore $\lim_{x \rightarrow x_k} f(x) = y_k = f(x_k)$.
