Let $U$ be an open set of $\mathbb{R}^n$, is it true that a closed set of $L^p_k(U)$ (with $p >1$, $k\geq 0 $) is closed in $L^2(U)$? (assuming it is contained in $L^2(U)$)
Is this true if $U$ is instead a compact subset of $\mathbb{R}^n$? and if $U$ is compact and $p=2$?
The answer is no. One example was already given: for $k\ge 1$, the Sobolev space $L_k^p$ is a closed subset of itself, but is a dense subset of $L^2$, because it contains smooth functions, and smooth functions are dense in $L^2$.
An example of different nature is given in $L^p_1(0, 1)$ by the sequence $$ f_n(x) = \begin{cases} 1-nx,\quad &0\le x\le \frac{1}{n} \\ 0, \quad & x>\frac1n \end{cases} $$ This sequence has no limit points in $L^p_1$ norm, and therefore $\{f_n\}$ is a closed set. But it is not a closed set in any $L^p$ for any $1\le p<\infty$ since $f_n\to 0$ in $L^p$ norm.