smallest filter Below in the block quotes is what I am trying to prove. Below the block quotes is my attempt for the first part of part (b). I guess I don't see what makes part a and part b different. For part a could I just change the way I define the filter?

Let $I$ be a nonempty set and let $\mathscr{C}$ be a nonempty collection of subsets of $I$.
(a) Prove that if $\mathscr{C}$ is closed under finite intersections and $\emptyset \notin \mathscr{C}$, then there exists $\mathscr{F} \supseteq \mathscr{C}$ such that $\mathscr{F}$ is the smallest filter containing $\mathscr{C}$.
(b) Prove that if $\mathscr{C}$ has the finite intersection property, i.e., if any finite intersection of elements of $\mathscr{C}$ is nonempty, then there exists $\mathscr{F} \supseteq \mathscr{C}$ such that $\mathscr{F}$ is the smallest
filter containing $\mathscr{C}$.

Proof (b): Define a filter $\mathscr{F}$ to be, $$\mathscr{F}=\{F : A_{1} \cap \dots \cap A_{n} \subset F \text{ for some } n \in \mathbb{N} \text{ and } A_{i} \in \mathcal{C}\}$$
First we want to show that the empty set is not in $\mathscr{F}$.
Suppose the empty set is contained in $\mathscr{F}$ such that for some $j \in \mathbb{N}$, $A_{1} \cap \dots \cap A_{j} \subseteq \emptyset$ which implies $A_{1} \cap \dots \cap A_{j} = \emptyset$. Hence a contradiction.
So $\emptyset \notin \mathscr{F}$
Now we want to show that if $A$ is an element of the filter and $A$ is a subset of $B$, then $B$ is also an element of the filter.
Let $A \in \mathscr{F}$ with $A$ a subset of $B$. Since $A \in \mathscr{F}$, there exists an $m \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{m} \subseteq A$, but $A \subseteq B$, so $A_{1} \cap \dots \cap A_{m} \subseteq B$.
Therefore B is an element of $\mathscr{F}$.
Now we will show that if A and B are elements of the filter, then $A \cap B$ is an element of the filter.
Let $A, B \in \mathscr{F}$. Then there exists an $n_{1} \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{n_{1}} \subseteq A$. And there exists and $n_{2} \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{n_{2}} \subseteq B$.
Let $k=max\{n_{1}, n_{2}\}$ so that $$A_{1} \cap \dots \cap A_{k} \subseteq A_{1} \cap \dots \cap A_{n_{1}} \subseteq A$$ and $$A_{1} \cap \dots \cap A_{k} \subseteq A_{1} \cap \dots \cap A_{n_{2}} \subseteq B$$
Hence $A_{1} \cap \dots \cap A_{k} \subseteq A \cap B$.
Therefore $A \cap B \in \mathscr{F}$.
Lastly we will show that $\mathcal{C}$ is contained in the filter, $\mathscr{F}$.
Let $A_{j}$ be any element of $\mathcal{C}$. Then we have that $A_{1} \cap A_{j}$ is an element of $\mathcal{C}$ and $A_{1} \cap A_{j} \cap A_{2} \subseteq A_{j}$. Thus $A_{j} \in \mathscr{F}$.
Hence, $A_{j}$ being an element of $\mathcal{C}$ implies $A_{j}$ is an element of $\mathscr{F}$ and thus $$\mathcal{C} \subseteq \mathscr{F}$$
Therefore $\mathscr{F}$ is a filter on $I$ with $\mathcal{C} \subseteq \mathscr{F}$.
 A: The difference between (a) and (b) is that in (a) we assume that $\bigcap\mathscr{B}$ is actually a member of $\mathscr{C}$ for each finite $\mathscr{B}\subseteq\mathscr{C}$, while in (b) we assume only that these intersections are non-empty. The hypothesis of (a) implies that of (b), so once you’ve proved (b), you’ve also proved (a).
Your argument showing that $\mathscr{F}$ is a filter is basically correct but does have a couple of flaws. You write:

Let $A, B \in \mathscr{F}$. Then there exists an $n_{1} \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{n_{1}} \subseteq A$. And there exists and $n_{2} \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{n_{2}} \subseteq B$.
Let $k=max\{n_{1}, n_{2}\}$ so that $$A_{1} \cap \dots \cap A_{k} \subseteq A_{1} \cap \dots \cap A_{n_{1}} \subseteq A$$ and $$A_{1} \cap \dots \cap A_{k} \subseteq A_{1} \cap \dots \cap A_{n_{2}} \subseteq B$$
Hence $A_{1} \cap \dots \cap A_{k} \subseteq A \cap B$.

First, you haven’t said what the sets $A_k$ are: it is important that they are elements of $\mathscr{C}$. That could be simply an oversight, but you also appear to be assuming that one of the collections $\{A_k:k=1,\ldots,n_1\}$ and $\{A_k:k=1,\ldots,n_2\}$ is a subset of the other. It is possible to show that they can be chosen in such a way that this is true, but it is not necessary to do so, and in any case you’ve not done so. I would just say that there are finite $\mathscr{C}_A,\mathscr{C}_B\subseteq\mathscr{C}$ such that $\bigcap\mathscr{C}_A\subseteq A$ and $\bigcap\mathscr{C}_B\subseteq B$. Then $\bigcap(\mathscr{C}_A\cup\mathscr{C}_B)\subseteq A\cap B$, and $\mathscr{C}_A\cup\mathscr{C}_B$ is a finite subset of $\mathscr{C}$, so $A\cap B\in\mathscr{F}$.
Finally, you worked much too hard in showing that each $\mathscr{C}\subseteq\mathscr{F}$. All that’s needed is that for each $A\in\mathscr{C}$, $\{A\}$ is a finite subset of $\mathscr{C}$, so $A=\bigcup\{A\}\in\mathscr{F}$.
To complete (b), just show that if $\mathscr{G}$ is a filter on $I$ such that $\mathscr{C}\subseteq\mathscr{G}$, then $\mathscr{F}\subseteq\mathscr{G}$.
