= Solution
By the <mean value theorem>, a <continuously differentiable function> with zero <derivative> on $[-1,1]$ is constant. Thus the <kernel> of the <differentiation operator> is
$$
\boxed{W=\{x\mapsto a:a\in\mathbb R\}=\langle1\rangle.}
$$
It is a <closed set> by the <finite-dimensional subspace is closed> theorem, applied in the <normed space> $(C^1([-1,1]),\|\cdot\|_\infty)$. It is proper because the <function> $x\mapsto x$ has nonzero <derivative>.
For $n\geq1$, take $u_n(x)=\sin(nx)$. Its <supremum norm> is at most one, whereas $u_n'(0)=n$ and $|u_n'(x)|\leq n$. Hence
$$
\boxed{\|Tu_n\|_\infty=n,\quad\|u_n\|_\infty\leq1;\qquad T\text{ is unbounded}.}
$$
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