Let $\Omega \subseteq \mathbb{R}^n$ be an open bounded domain. Suppose that $f \in W^{1,p}(\Omega)$ for some specific $1<p<n$, and that $f$ is continuous.
Is it true that $f \in W^{1,n}_{loc}(\Omega)$?
(In "almost" the other direction, we know that if $f \in W^{1,p}$ for $p>n$ then it's continuous).
No, there is no such self-improvement. Continuity adds nothing for Sobolev regularity purposes.
Consider functions of the form $u(x)=|x|\sin (|x|^{-a})$, $u(0)=0$, where $a>0$. The domain can be the unit ball $B$.
The function $u$ is continuous on $B$ and smooth on $B\setminus\{0\}$. The gradient of $u$ is $$ \nabla u(x) = \frac{x}{|x|}\sin (|x|^{-a}) - \frac{a x}{|x|^{1+a}}\cos (|x|^{-a}) $$ which implies $|\nabla u(x)| = O(|x|^{-a})$, and $|\nabla u(x)| \sim |x|^{-a}$ most of the time, when the cosine term is bounded away from $0$.
Therefore, $u\in W^{1,p}(B)$ if and only if $p<n/a$. Choose $a>0$ as you wish.