Checking if a set makes up a filter 
Definition: A non-empty family $\Lambda$ of subsets of a non-empty set $X$ is called
a filter on $X$ if it satisfies the following three conditions:


(i) $\emptyset \not \in  \Lambda$,


(ii) $A, B \in \Lambda \implies  A \cap B \in \Lambda$,


(iii) $(A \in \Lambda, A \subseteq S \subseteq X) \implies S \in \Lambda$.



*

*Consider $X = \{1, 2, 3\}$. Then $\Lambda = \{\{1\}, \{1, 2\}, \{1, 2, 3\}\}$ is a filter on $X$ because
(i) $\emptyset \not \in \Lambda$
(ii) $\{1\} \cap \{1, 2\}, \ \{1\} \cap \{1, 2, 3\}, \ \{1, 2\} \cap \{1, 2, 3\} \in \Lambda$
(iii) $\{1\} \subseteq \{1\} \subseteq \{1, 2\} \subseteq \{1, 2, 3\} \subseteq \{1, 2, 3\} = X$, but $\{1\}, \{1, 2\}, \{1, 2, 3\} \in \Lambda$


*Let $\mathcal N(x)$ be the family of neighborhoods of a chosen point $x$ in a topological
space $(X, \tau)$. Then  $\mathcal N(x)$ is a filter on $X$ because
(i) $x \in N_i \in \mathcal N(x) \text{ for all $i$}  \implies \emptyset \not \in \mathcal N(x)$
(ii) $x \in N_i \in \mathcal N(x) \text{ for all $i$} \implies x \in \bigcap N_i \implies \bigcap N_i$ is a neighborhood of $x$ and so $\bigcap N_i \in \mathcal N$
(iii) $N_i \subseteq N_j \subseteq X \implies N_j \in \mathcal N(x)$ by definition


*$\{D_i\}_{i=1}^\infty (\subseteq \mathbb R)$ with $(-\infty, a] \subseteq D$ for some $a$ is a filter on $\mathbb R$ because
(i) $(-\infty, a] \neq \{\} \implies D_i \neq \{\} \implies \{\} \not \in \{D_i\}_{i=1}^\infty $
(ii) $(\infty, x] \cap (-\infty, y] = (\infty, x] \iff x < y$ and $(\infty, x] \cap (-\infty, y] = (\infty, y] \iff x \ge y$; either intersection is a subset of $D \in \{D_i\}_{i=1}^\infty$
(iii) $D_1 \subseteq D_2\subseteq \mathbb R \implies D_2 \in \{D_i\}$ by definition


*$T = \tau_{cocountable} \setminus \{\emptyset\}$ is a filter on any uncountable set because
(i) $\emptyset \not \in T$
(ii) Suppose $A, B \in T$. Then $A^c, B^c$ are countable and so  $A^c \cup B^c = (A \cap B)^c$ is countable and so $A \cap B \in T$
(iii) this criterion follows by definition

Are the proofs above convincing?
 A: All the ones except the third one looks good.  The third one requires an additional criteria that there is no upper bound on the chosen $a$'s that make up the $[-\infty,a]$ sets in your filter.  If there was an upper bound,  then you would no longer be closed under supersets, as adding things beyond that upper bound would no longer be in your filter
A: You only have to consider the intersection of two sets. For the third you seem overly ambitious ( $\{D_i\}_{i=1}^\infty$ ?). Just say
(ii): if $A,B$ are in the filter, for some $x,y \in \Bbb R$ we have $(-\infty,x] \subseteq A, (-\infty, y] \subseteq B$. Then
$$ (-\infty,x] \cap (-\infty,y] = (-\infty,\min(x,y)] \subseteq A \cap B$$
so that $A\cap B$ is in the filter too.
For the last, (iii) I would still mention that $A \subseteq B$ implies $B^\complement \subseteq A^\complement$ and that a subset of a countable set is still countable. Don't chicken out with "by definition" when it can easily be explained by a single line.
Where do you use the finiteness of the intersection in the second example? What's the relation with the topology?
