Prove that the following statements are equivalent if $\tau_1$ is finer than $\tau_2$. Setup for question:

Let $A$ bet a set. Let $\tau_2$ and $\tau_1$ be two topologies on $A$.

Statements to prove that are equivalent:
(I) $\tau_2 \subseteq \tau_1$ (i.e. $\tau_1$ is finer than $\tau_2$);
(II) for every $a\in A$ and $U \in \tau_2$ with $a\in U$ there exists $V \in \tau_1$ such that $a \in V \subseteq U$; and
(III) for every $a\in A$ and $U \in \tau_2$ with $a\in U$ there is a finite set $F \subseteq \tau_1$ such that $a \in \bigcap F \subseteq U$.
WORKING
(I) $\implies$ (II):
Suppose that $\tau_1$ is finer than $\tau_2$, i.e. $\tau_2 \subseteq \tau_1$. Let $a \in A$ and let $U \in \tau_2$. Then $U\in \tau_1$. If we set $U=V$ then $a\in V \subseteq U$.
(II) $\implies$ (I)
Suppose that for every $a\in A$ and $U \in \tau_2$ with $a\in U$ there exists $V \in \tau_1$ such that $a \in V \subseteq U$. 
Let $B \in \tau_2$. For each point $a \in B$ there's a guaranteed existence of a set $S_a \in \tau_1$ such that $a \in S_a \subseteq B$. Indeed:
$$ B = \bigcup_{a \in B}S_a.$$
Since arbritary unions of $\tau_1$-open sets are $\tau_1$-open, the above equality shows that $B$ is $\tau_1$ open, i.e. $ B \in \tau_1$. Therefore, $\tau_2 \subseteq \tau_1$.
Help needed for (I) $\iff$ (III) OR (II) $\iff$ (III).
 A: Your proof of $(I) \iff (II)$ is correct. To prove $(II) \implies (III)$ we can take $F = \{V\}$. Then $F$ is clearly a finite subset of $\tau_1$, and $$a \in \bigcap F = \bigcap \{V\} = V \subseteq U\,.$$
For the other direction: By definition of a topology, $V := \bigcap F$ is a member of $\tau_1$, and by assumption on $F$ it satisfies the condition in $(II)$.
A: We show:
$I)\Rightarrow II)\Rightarrow III)\Rightarrow I)$
$I)\Rightarrow II):$
Let $a\in A$ be arbitrary and $U\in\tau_2$ with $a\in U$. Since $\tau_2\subseteq \tau_1$ it is $U\in\tau_1$ and we can choose $V=U\ni a$.
$II)\Rightarrow III):$
Let $a\in A$ be arbitrary and $U\in\tau_2$ with $a\in U$.
We have to show, that it exists a finite set $\mathcal{F}\subseteq\tau_1$, such that $a\in\bigcap_{F\in\mathcal{F}} F\subseteq U$.
We know from the assumption, that it exists $V\in\tau_1$ with $a\in V\subseteq U$.
Choose $\mathcal{F}=\{V\}$, which is a finite set and $\bigcap_{F\in\{V\}} F=V\subseteq U$.
$III)\Rightarrow I):$
Let $U\in\tau_2$. We have to show, that $U\in\tau_1$.
Let $a\in U$ be arbitrary. Then exists a finite set $\mathcal{F}_a\subseteq\tau_1$ with $a\in\underbrace{\bigcap_{F\in\mathcal{F}_a}}_{\in\tau_1} F\subseteq U$.
[It is $\bigcap_{F\in\mathcal{F}_a} F\in\tau_1$ because of the definition of a topology. Which states, that the intersection of finite open sets in $\tau_1$ is again in $\tau_1$ and $\mathcal{F}_a$ just contains finite elements, so we take a finite interesection!]
Then is $\underbrace{\bigcup_{a\in U} \mathcal{F}_a}_{\in\tau_1}=U\in\tau_1$
[It is $\underbrace{\bigcup_{a\in U} \mathcal{F}_a}\in\tau_1$, because we know $\mathcal{F}_a\in\tau_1$ for every $a\in U$. From the definition of a topology we know, that an arbitrary union of sets in $\tau_1$ is again in $\tau_1$!]
Which concludes the proof.
