= Solution
The <intermediate value theorem> states that a continuous real-valued function on $[a,b]$ takes every value between $f(a)$ and $f(b)$. To prove it, suppose $f(a)<k<f(b)$ and repeatedly bisect the interval, keeping endpoints whose function values bracket $k$. The resulting nested closed intervals have lengths tending to zero and a common point $c$ by <completeness of the real numbers>. Both endpoint sequences tend to $c$, so <continuity> implies $f(c)=k$. Endpoint values are immediate, and reversing the inequalities handles $f(a)>f(b)$.
For the derivative assertion, put $m=(f(b)-f(a))/(b-a)$. The <intermediate secant slope from endpoint derivatives> can be constructed in three cases. If $m=k$, take $[a',b']=[a,b]$. If $m>k$, the function
$$
g(x)=\frac{f(x)-f(a)}{x-a},\qquad g(a)=f'(a),
$$
is continuous on $[a,b]$: differentiability at $a$ supplies the endpoint limit, and differentiability elsewhere implies continuity. Since $g(a)<k<g(b)$, the <intermediate value theorem> gives $x\in(a,b)$ with $g(x)=k$. Take $[a',b']=[a,x]$.
If $m<k$, use instead
$$
h(x)=\frac{f(b)-f(x)}{b-x},\qquad h(b)=f'(b).
$$
Its continuous endpoint extension satisfies $h(a)<k<h(b)$, so there is $y\in(a,b)$ with $h(y)=k$. Take $[a',b']=[y,b]$. In every case,
$$
\frac{f(b')-f(a')}{b'-a'}=k.
$$
The <mean value theorem> on this subinterval now gives
$$
\boxed{\text{some }c\in(a,b)\text{ satisfies }f'(c)=k.}
$$
This proves the <Darboux theorem for derivatives> without assuming that $f'$ is continuous.
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