Assume It is known know that oscilation $\omega(f_n, U) > \epsilon$ for continuous functions $f_n$ and some open set $U$ starting from some $N$. Also it is known that $f_n \to \varphi$ pointwise. It is also safe to assume that $\overline{U}$ is compact..
Here $$\omega(f,U) = \sup_{x,y \in U} \Big|f(x) - f(y)\Big|$$
Intuitively, it seems true that that $\omega(\varphi,U) \ge \epsilon$.
But is it actually true?
I thought that it must be possibele to find points $x_n,y_n \in \overline{U}$ such, that
$$
\omega(f_n,U) \le \Big|f_n(x_n) - f_n(y_n) \Big|
$$
Then compute limits of converging subsequences, which exist by compactness of $\overline{U}$. Name them $Y$ and $X$.
And Then to show that $|\varphi(Y) - \varphi(X)|$ is substantially big by convergence and continuity arguments. But I don't know how to do it as both argument and function depend on $n$.
To answer the question as is: No. We do not necessarily have that $\omega(\varphi,U) \ge \epsilon$.
For example, take
$$f_n : (0,1) \rightarrow \mathbb{R} : x \mapsto \frac{1}{nx}$$
Then, for all $n$, we have that $\, \omega(f_n, U) \ge \epsilon \,$, but since $\, f_n \xrightarrow{\text{p.w.}} 0 \,$, $\,\omega(\varphi,U) = 0$.
To me it seems that the more interesting question would be to replace $U$ with $\overline{U}$.