Why is the derivative operator $D:C^\infty([0,1])\to C([0,1])$, $Df\equiv f'$, closed and unbounded?

This is an example I read around closed graph theorem. Let $Y=C[0, 1]$ and $X$ be its subset $C^\infty[0, 1]$. Equip both with uniform norm. Define $D: X \to Y$ by $f \mapsto f'$. Suppose $(f_n, f_n') \to (g, h)$ in $X \times Y$. Thus, $f_n \to g, f_n' \to h$ both uniformly on $[0, 1]$. It is well known that $g' = h$. Thus, the graph of $D$ is closed. But since $D(t^n) = nt^{n-1}$, $D$ is not continuous. I have three questions about this example.

1. Why $f_n \to g, f_n' \to h$ both uniformly on $[0, 1]$?
2. Why $g' = h$ even though it is well known?
3. Why $D$ is not continuous because $D(t^n) = nt^{n-1}$?

Could anyone explain these questions to me, please? Thank you!