Why does a filter base uniquely define a filter?
We define the filter base $\mathcal B$ of the filter $\mathcal F$as:
$\forall F \in \mathcal F \ \exists \ B \in \mathcal B: B\subset F$
So why can $\mathcal B$ be the filter base of at most one filter?
Let $\mathcal{F}_1$ and $\mathcal{F}_2$ be two filters that have $\mathcal{B}$ as a base.
Recall again that $\mathcal{F}$ has $\mathcal{B}$ as base iff $\mathcal{B} \subseteq \mathcal{F}$ and $\forall F \in \mathcal{F} \exists B \in \mathcal{B} : B \subseteq F$
Then $F \in \mathcal{F_1}$ means there exists a $B \in \mathcal{B}$ such that $B \subseteq F$. But as $\mathcal{B} \subseteq \mathcal{F_2}$ as well, the filter axioms imply $F \in \mathcal{F}_2$ as well. This shows $\mathcal{F}_1 \subseteq \mathcal{F}_2$ and the reverse inclusion is proved symmetrically. Hence equality ensues.