Chang & Keisler Exercise 4.3.9 I'm having a lot of trouble sorting out an exercise from Chang & Keisler's Model Theory:

4.3.9: Let $I$ be an infinite set of power $\alpha$. If $E \subset P(I)$, $|E| \leq \alpha$, and the filter generated by $E$ is uniform, then $E$ can be extended to an $\alpha$-regular ultrafilter $D$ over $I$.

In chapter 4.3, CK proves that for any infinite $\alpha$, there exists an $\alpha$-regular ultrafilter on $\alpha$. Their construction involves working with the set of finite subsets of $\alpha$ and I tried mimicking that argument, but I'm not sure how to employ the given hypotheses. Any and all hints / suggestions are appreciated.
 A: A modification of the proof used in Chang and Keisler can be used to build a regularizing family $E_i$ which is compatible (ie has the finite intersection property) with the uniform filter created by extending $E$ as given in the hypothesis of the problem. Any extension of an $\alpha$-regular filter is $\alpha$-regular so this concludes to problem. Uniformity is used to ensure that the regularizing family is indeed compatible since otherwise the construction would fail. This can also be seen by the fact that every regular filter is uniform. Here is my proof:
Denote the set of finite subsets of $I$ by $P_\omega(I)$. Let $f:I\rightarrow{} S_\omega(I)$ be an arbitrary surjection. For all elements $i\in I$ define the sets
$E_i=\{j\in I: i\in f(j)\}$
Now it is clear that the family of sets $\{E_i\}$ for all $i\in I$ has the finite intersection property since
$E_i\cap E_k=\{j\in I: i\in f(j)\wedge k\in f(j)\}\neq \emptyset$
because there exists a finite sequence in $P_\omega(I)$ which contains any pair $i,k$.
Now if we consider the uniform filter $E'$ generated by $E$ given at the start of the problem, we know that $E'$ has the finite intersection property by way of already being a filter. If we can show that $E'\cup\{E_i\}$ has the finite intersection property then the union can be extended into a regular ultrafilter over $I$ by the ultrafilter lemma.
Since $E'$ and $\{E_i\}$ have the finite intersection property within their respective families it remains to be shown that any intersection between $A\in E'$ and $E_i$ for $i\in I$ is nonempty for arbitrary $A\in E'$ and $i\in I$.
Before showing this intersection is non empty recall that the function $f$ which defined the $E_i$ was arbitrary. We must now define it so that the finite intersection property holds.
Enumerate $A\in E'$ as $A_{\beta}, \beta<\alpha$ as well as every element within $A_\beta$ as $\lambda_\beta<\alpha$.
Likewise enumerate the elements of $P_\omega(I)$ as $\gamma$ with $\gamma<\alpha$. Now we build $f$ inductively by starting with a series of partial functions $g_i$ which are then extended to partial functions $f_i$ which is extended to $f$.
Let $g_{0_0}$ map $0_0\in A_0$ to $\gamma=0$. Let $g_{0_1}$ map $0_1\in A_1$ to $\gamma=0$ if $A_0$ does not contain $\gamma=0$ . Likewise let $g_{0_\beta}$ map $0_\beta\in A_\beta$  to $\gamma=0$ if $_\beta$ does not contain $0$ otherwise do nothing. Let $f_0=\cup g_0$. Now repeat the procedure with $g_{1_\beta}$ taking the union to get $f_1$ and so on to get $f_i$ finally get $f$ which is then defined upon every element in $I$. The reason this can be done is that $E'$ is uniform and that every set $A_\beta$ is of size $\alpha$. This ensures that there is always "room" to extend the function since otherwise would amount to an element in the filter of size less than $\alpha$. $f$ has the special property that every set $A_0$ is surjective onto the domain $P_\omega(I)$. Thus the for any set $A\in E'$ and $E_i$, $A\cap E_i$ is nonempty. Therefore as desired the family of sets $E'\cup \bigcup_{i\in I }E_i$ has the finite intersection property and a regularizing family of size $\alpha$. This ensures that it can be extended to a $\alpha$-regular ultrafilter as desired.
