$\psi:\Omega\to\mathbb{R}$ is continuous. $\Omega$ is a bounded open subset of $\mathbb{R}^m$. $f_n\in H^k(\Omega)$, $f_n$ is continuous and $k>\frac{m}{2}$. It is given that
$$\limsup\limits_{n\to\infty}\left|f_n(x)-\psi(x)\right| = \epsilon\forall x\in D,$$ $D$ is a countable dense subset of $\Omega$.
It is known that $$\lim\limits_{n\to\infty} \sum_{|\alpha|=k}\|D^{\alpha}f_n \|^2_{L^2(\Omega)} = O(1)$$ As $k>\frac{m}{2}$ what the above equation says is the sequence $\{f_n\}$ does not converge to a discontinuous function.
Can we say that there exists a $\delta(\epsilon)$ such that $$\limsup_{n\to\infty}\|f_n(x)-\psi(x)\|_{L^{\infty}(\Omega)} = \delta(\epsilon) $$
By Morrey's inequality, you have that the functions are uniformly $\alpha$-Holder continuous for some $\alpha>0$. Then you pick a finite set of points in $D$ so that every point in $\Omega$ is within distance $\eta>0$ of one of these finite points, where $\eta$ is sufficiently small. Then you find $N$ such that for $n≥N$, $|f_n(x)−ψ(x)|<2ϵ$ for all $x$ in that finite set. Then you use the uniform $α$-Holder nature of the $f_n$ to get the bound you want. And $δ(ϵ)=3ϵ$ if you choose $η$ small enough. (You could actually get $\delta(\epsilon) = \epsilon$ if you push it a little more.)