real analysis - Is intermediate value property equivalent to Darboux property?

I always thought that a function $f:\mathbb{R} \to \mathbb{R}$ has the intermediate value property (IVP) iff it maps every interval to an interval (Darboux property):

Proof:

Let $f$ have the Darboux property and let $a Then $f([a,b])$ is an interval and so contains $[f(a),f(b)]$. If $u
> \in [f(a),f(b)]$ then of course $u \in f([a,b])$ and thus there exists
some $k \in [a,b]$ such that $f(k) = u$, i.e. $f$ has IVP.

Now let $f$ have the IVP and let $a < b$, $x,y \in f([a,b])$ and $z

> \in \mathbb{R}$ with $x have $x = f(x')$ and $y = f(y')$ for some $x',y' \in [a,b]$. W.L.O.G
assume that $x' < y'$. Then by the IVP there is some $c\in [x', y']$
such that $f(c) = z$, i.e. $z \in f([a,b])$ and therefore $f([a,b])$
is an interval.

But on this blog in problem 5. the author says that they are not equivalent:

"This [Darboux property] is slightly different from continuity and
intermediate value property. Cotinuity implies Darboux and Darboux
implies Intermediate value property."

Have I missed something and if yes, where does the proof given above break?

Answer

That is correct according to the definition of intermediate value property saying that for all $a

The blog's author Beni Bogoşel clarified in a comment that he was using a different definition of intermediate value property, meaning that the entire image is an interval. The ambiguity is understandable given that the Intermediate Value Theorem is often stated in terms of a particular interval: $f:[a,b]\to \mathbb R$ continuous implies $f([a,b])$ contains the interval $\{\min\{f(a),f(b)\},\max\{f(a),f(b)\}\}$. (In this case, because the restriction of a continuous function is also continuous, this theorem automatically implies the stronger intermediate value property for continuous functions on intervals.)

The author acknowledged that the convention is not universal, and he may edit to clarify.