As the title suggests, I would like to know whether the set
$$\{ f\in W^{m, 2}(0,1): \|f-\psi\|_{\infty} \leq 1 \}$$
is sequentially compact in $W^{m,2}(0,1)$. Here, $W^{m,2}(0,1)$ denotes the Sobolev space of functions defined on $(0,1)$ that have square integrable $m$th derivatives and norm
$$\|f\|_{W^{m,2}} = \sqrt{\|f\|^2_{L^2} + \|f^{(m)}\|_{L^2}^2}$$
and $\|f\|_{\infty}$ stands for the $\sup_{x \in (0,1)} |f(x)|$, the uniform norm. I believe that the set is indeed sequentially compact and this can be shown by an application of the Arzela-Ascoli theorem. Is this correct?
No, this set is not compact in $W^{m,2}$. It is not even bounded. Take $\psi=0$, $$ v_n(x):=\sin(n\pi x). $$ Then the $v_n$ are in the set, but $\|v_n \|_{H^m}\to\infty$ for all $m\ge1$.