Fix some irrational $\alpha$ and any enumeration $\{q_n:n\in\Bbb Z^+\}$ of the rationals. We’ll build a new enumeration $\{p_n:n\in\Bbb Z^+\}$ of $\Bbb Q$ in such a way that for each $n\in\Bbb Z^+$, $\alpha\notin(p_n-\frac1n,p_n+\frac1n)$.
Let $n_1=\min\{n\in\Bbb Z^+:|q_n-\alpha|\ge 1\}$, and let $p_1=q_{n_1}$; clearly $\alpha\notin(p_1-1,p_1+1)$. Let $Z_1=\Bbb Z^+\setminus\{n_1\}$, the set of indices of rationals not yet re-enumerated.
Now suppose that we’ve already defined $p_1,\dots,p_m$ and $Z_m$. Let $$n_{m+1}=\min\left\{n\in Z_m:|q_n-\alpha|\ge\frac1{m+1}\right\}\;,$$ and set $p_{m+1}=q_{n_{m+1}}$ and $Z_{m+1}=Z_m\setminus\{n_{m+1}\}$. Clearly $p_{m+1}$ is distinct from $p_1,\dots,p_m$ and $$\alpha\notin\left(p_n-\frac1n,\,p_n+\frac1n\right)\;.$$
All that’s left is to prove that every rational is eventually enumerated as $p_n$ for some $n\in\Bbb Z^+$. That follows from the fact that at each stage we took the first available rational in the original enumeration; I’ll leave it to you to fill in the details, unless you get stuck and ask for help.
Uhm, I only have a suggestion for part a. Notice that $E_{\epsilon,\delta}$ almost look like the inverse image of an open set. I know, $\leq$ does not make it look like the inverse image of an open set. But if you divide the set in $E^1$ and $E^2$, where for $E^1$ the strict inequality holds, and for $E^2$ equality holds, then $E^1\cap E^2=\emptyset$, and $E^1$ IS the inverse image of an open set. As for $E^2$, well, can there actually be any element in there?
Edit: let me explain better.
1) If $E^2$ is not empty, it means $\exists x$ such that $|f(x,y)-f(x,0)|=\epsilon, \forall y<\delta$. But if $f(x,\cdot)$ is continuous, this cannot happen.
2) This is more complicate than I thought. And David's hint is really important. Define $F_y(x)=f(x,y)-f(x,0)$. Then $A_y = F_y^{-1}(-\epsilon,\epsilon)$ is measurable. As David says, the definition of $E_{\epsilon,\delta}$ can be replaced by
$$
E_{\epsilon,\delta}=\{x:|F_y(x)|<\epsilon,\forall y<\delta, y\in\mathbb{Q}\}
$$
Therefore, it follows that $E_{\epsilon,\delta}=\cap A_y$, where the intersection is over all $y<\delta,y\in\mathbb{Q}$. Since each $A_y$ is measurable and the intersection is countable, we obtain that $E_{\epsilon,\delta}$ is measurable.
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