Let $\mathbb{A} \subset P(A)$ non-empty and $\mathbb{A}$ have finite intersection property, Let $\mathbb{A} \subset F,F' $ such that $F,F'$ are filters and $F = \{ X : \exists T (T \subset \mathbb{A} , |T| < \aleph_0 , \bigcap T \subset X \subset A) \}$ and the question asks to prove that $F \subset F'$.
I assumes that $X \in F$ but $X \not\in F'$ from the fact that $X \in F$ , there is $T \subset \mathbb{A}$ such that $|T| < \aleph_0$ and $\bigcap T \subset X$
From here I want to arrive to contradiction, but I don't know how?
Note : $F$ is a filter means that :
1) $F$ is not empty and $A \in F$ and $F \subset P(A)$
2) $F$ have the finite intersection property meaning for all $B \subset F$ that is not empty we have that $\bigcap B \not= \emptyset$
3) If $B \in F$ and $B \subset C \subset A$ then $C \in F$
We prove $F \subset F'$. Suppose $X \in F$ so we have some finite $T = \{ Y_1, ..., Y_n\} \subset \mathbb{A}$ such that $\cap T = Y_1 \cap ... \cap Y_n \subset X$. Ok now lets look at $F'$. $\mathbb{A} \subset F'$ so we have $Y_i \in F'$ for $i \in \{1, ..., n\}$. Now because $F'$ is a filter we can easily see that $Y_1 \cap ... \cap Y_n \in F'$ and again since $F'$ is a filter, it should be upward closed and must contain all subsets of $A$ that contain $Y_1 \cap ... \cap Y_n$, and since $X$ contains this too, we conclude that $X \in F'$.