$D(0) =\{f:[a,b]\to\Bbb{R}: f \text{ has the Darboux Property}\}$
$D(1) =\{f:[a,b]\to\Bbb{R}: \exists (f_n)\subset D(0)\text{ and } f_n \overset{\text{pointwise}}\to f\}$
$B(0) =\{f:[a,b]\to\Bbb{R}: f \text{ is continuous}\}$
$B(1) =\{f:[a,b]\to\Bbb{R}: \exists (f_n)\subset B(0)\text{ and } f_n \overset{\text{pointwise}}\to f\}$
Then $B(0) \subset B(1) $ and $D(0) \subset D(1) $
Again $B(0) \subset D(0) \subset D(1) $
Question 1: Is $B(1) $ and $ D(0)$ comparable under set inclusion $\subset$?
Attempt:
$f:[0, 1]\to \Bbb{R}$ defined by $f(t)= \begin{cases}1& t\in (0,1]\\0& t=0\end{cases}$
Then $f\in B(1) $ but $f\notin D(0) $
$f\in D(0) $ i.e $f:[0, 1]\to\Bbb{R}$ has the Darboux property.
$f:[a, b]\to \Bbb{R}$ differentiable then $f'\in D(0) $ but $f'\in B(1) $ aslo i.e $\{f' :f\in D[a,b]\}\subset B(1) \cap D(0) $
Indicator function of $\Bbb{Q}$, the Cantor set are in $B(1) $ but none of the function have the Darboux property.
Finally I have found an example $($thanks to J.H.Conway$)$ Conway base 13 function which is in Baire class $2$ but have the Darboux property.
$\boxed{\text{Answer ( of Question $1$) : No.}}$
Question $2$ : Under what minimum assumption (weaker assumption) $D(0) \subset B(1) $ ?
Points to be considered :
• If $f\in D(0) $ and $f$ preserve compact sets then $f\in B(0) \subset B(1) $.
• If $f\in D(0) $ has a primitive then $f\in B(1) $
•$D_f$ is countable then $f\in B(1) $. Here $f$ is not an ordinary function,$f\in D(0) $.What kind of simplification it can bring in $D_f$ ?