Frm Folland's Real Analysis exercise 2.43 part (b):
Suppose that $(X, \mathcal M, \mu)$ is a measure space, with $\mu(X) < \infty$. Let $f: X × [0,1] \to \mathbb C$ is a function such that $f(\cdot, y)$ is measurable for each $y \in [0,1]$ and $f(x, \cdot)$ is continuous for each $x \in X$. Prove that: for any $\epsilon > 0$ there is $E \subset X$ such that $\mu(E) < \epsilon$ and $f(\cdot, y) \to f(\cdot, 0)$ uniformly on $E^c$ as $y \to 0$.
Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < \epsilon, \delta < 1$ then $E_{\epsilon, \delta} = \{x : |f(x, y) - f(x, 0)| \leq \epsilon \text{ for all } y < \delta \}$ is measurable. Again, I already proved this; no need for you to prove this.
I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).
Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that \begin{align*} \bigcap_{n\in \mathbb{N}} (E_{1/m,1/ n})^c = \emptyset \quad \forall m\in \mathbb{N}. \end{align*} Let $A_{m,k}=\bigcap_{n\leq k} (E_{1/m,1/ n})^c$. For each $m\in \mathbb{N}$, by upper continuity, there is some $k\in \mathbb{N} $ for which \begin{align*} \mu(A_{m,k})<\frac{\epsilon}{2^m}. \end{align*} Thus, \begin{align} E:=\bigcup_{m\in \mathbb{N}} A_{m,k} \end{align} has measure less than $\epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.