By definition, if the derivative operator $D:C^1[-1,1]\to C^1[-1,1]$ is closed, then it should be the case that, given any sequence $\{x_n\}$ in $C^1[-1,1]$, and given that $x_n\to x$ as $n\to\infty$ for some $x\in C[-1,1]$, we should conclude that $x$ is in $C^1[-1,1]$. But it seems that this is not the case.
Consider $x_n(t)=\sqrt{n^{-2}+x^2}$, clearly $x_n$ are in $C^1[-1,1]$, and we know that $x_n\to x(t)=|x|$, which is not in $C^1[-1,1]$, and doesn't it show that $D$ is not a closed linear operator?