 Solution

Differentiation satisfies $\Vert Tf{\Vert }_{\infty }\le \Vert f\Vert$, so its operator norm is at most $1$. With $f_n(t)=\sin (2\pi nt)/(2\pi n)$, the ratio $\Vert T{f}_n{\Vert }_{\infty }/\Vert f_n\Vert$ tends to $1$. Hence $\Vert T\Vert =1$.

Normalize $g_n=f_n/(1+(2\pi n{)}^{-1})$, so $\Vert g_n\Vert =1$. Their derivatives are $\cos (2\pi nt)/(1+(2\pi n{)}^{-1})$. Distinct cosine functions are orthogonal in $L^2(0,1)$, and these normalizing factors stay bounded away from zero. Consequently the derivative images have pairwise $L^2$ distances bounded away from zero. Since uniform convergence would imply $L^2$ convergence, no subsequence converges uniformly. Thus $T$ is not compact.