Does Darboux property imply strong Darboux property? Given a set $S$, we say that a monotone set function $f\colon \mathcal{P}(S) \to [0,1]$ has the strong Darboux property whenever $f(\emptyset)=0$, $f(S)=1$, and for all $A\subseteq B \subseteq S$ and $x \in [f(A),f(B)]$ there exists $A\subseteq X\subseteq B$ for which $f(X)=x$.
If this property holds only for $A=\emptyset$, then $f$ is said to have the Darboux property.
Question: Does there exist a monotone set function $f$ which has the Darboux property and it hasn't the strong Darboux property?
 A: Yes, it exists. Let $\mathsf d^\ast$ be the upper asymptotic density on $\mathbf N$, viz. the function $$\mathcal P(\mathbf N) \to \mathbf R: X \mapsto \limsup_{n \to \infty} \frac{|X \cap [1, n]|}{n},$$ and let $V \subseteq \mathbf N$ be such that ${\sf d}^\ast(V) = 0$ and $|V| = \infty$ (e.g., $V$ can be the set of all squares). Then consider the function
$$f: \mathcal P(\mathbf N) \to [0,1]:
X \mapsto \left\{
\begin{array}{ll}
\mathsf{d}^\ast(X) & \text{if }V \subseteq X \\ 0 & \text{otherwise}\end{array}\right..$$ 
It is relatively easy to see that $f$ is monotone: It is essentially enough to consider that $\mathsf d^\ast$ is monotone in its own right.
Yet, $f$ has not the strong Darboux property: If $X \subseteq \mathbf N$ is such that ${\sf d}^\ast(X) = 1$ and $V \not\subseteq X$ (e.g., we can set $X := \mathbf N \setminus V$), then $[f(X), f(\mathbf N)] = [0,1]$, but $f(Y) \in \{0, 1\}$ whenever $X \subseteq Y \subseteq \mathbf N$.
So we are left to show that $f$ has the Darboux property. But this is more or less straightforward: Fix $X \subseteq \mathbf N$ and let $\alpha \in [0, f(X)]$. If $f(X) = 0$, there is nothing to prove. Otherwise, $V \subseteq X$, and it follows from the fact that ${\sf d}^\ast$ has the Darboux property, see e.g. 

G. Grekos, Répartition des densités des sous-suites d'une suite
  d'entiers, J. Number Theory 10 (1978), No. 2, 177-191,

that there exists $Y \subseteq X$ such that ${\sf d}^\ast(Y) = \alpha$, with the result that $Y \cup V \subseteq X$ and $f(Y \cup V) = \alpha$.
A: Here is a better example, showing that not only the answer to the OP is positive, but it is even possible to enforce on $f$ some additional, and fairly strong, conditions.
To start with, let $\theta$ be the function $\mathcal P(\mathbf N^+) \to \mathbf R$ defined as follows: Given $X \subseteq {\bf N}^+$, we assume $\theta(X) := 0$ if $|X| < \infty$, otherwise $\theta(X) := \sup_{n \ge 1} (x_{n+1} - x_n)^{-1}$, where $(x_n)_{n \ge 1}$ is the natural enumeration of $X$. Next, let $f$ be the function
$$
\mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto \sqrt{\theta(X)\, {\sf d}^\ast(X)},
$$
where ${\sf d}^\ast$ is the upper asymptotic density on $\mathbf N^+$.
It is seen that $f$ is monotone and affinely $(-1)$-homogeneous, essentially because the same is true of $\theta$ and ${\sf d}^\ast$, see the note below; moreover, $f(\emptyset) = 0$ and $f(\mathbf N^+) = 1$. 
We want to show that $f$ has the weak Darboux property. For this, fix $X \subseteq \mathbf N^+$ with $f(X) > 0$. Then $\mathsf{d}^\ast(X) > 0$, which implies that $|X| = \infty$ and there exist $x,y \in X$ such that $x < y$ and $f(X) = (y-x)^{-\frac{1}{2}}\sqrt{\mathsf{d}^\ast(X)}$.
Accordingly, let $a \in {]0, f(X)]}$, so that $0 \le (y-x) a^2 \le \mathsf{d}^\ast(X)$.
Since $\mathsf{d}^\ast$ has the weak Darboux property, see [1], we can find $Y \subseteq X$ such that $\mathsf{d}^\ast(Y) = (y-x) a^2$; set $A := Y \cup \{x,y\}$. It then follows from the monotonicity of $\theta$ that $\theta(A) = \theta(X) = (y-x)^{-1}$, while it is clear that $\mathsf{d}^\ast(A) = \mathsf{d}^\ast(Y)$. Thus, it is seen that $f(A) = a$, which ultimately shows that $f$ has the weak Darboux property.
Now, let $X := 4 \cdot \mathbf N^+$ and $Y := X \cup \{2\}$. It is straightforward that $\theta(X) = \mathsf{d}^\ast(X) = \mathsf{d}^\ast(Y) = \frac{1}{4}$ and $\theta(Y) = \frac{1}{2}$, with the result that $f(X) = \frac{1}{4}$ and $f(Y) = \frac{1}{4}\sqrt{2}$. Therefore, $f$ does not have the strong Darboux property, because $f(X) < f(Y)$, but $Y \setminus X$ is finite.
Notes. Here, a function $g: \mathcal P(\mathbf N^+) \to \mathbf R$ is said to be affinely $(-1)$-homogeneous if $g(k \cdot X + h) = g(X)$ for all $X \subseteq \mathbf N^+$ and $h, k \in \mathbf N^+$, where $k \cdot X + h := \{kx+h: x \in X\}$.
Bibliography.
[1] G. Grekos, Répartition des densités des sous-suites d'une suite d'entiers, J. Number Theory 10 (1978), No. 2, 177-191
