Neighborhood base at zero in a topological vector space I'm reading  a proof of a theorem which characterizes the collections of sets which can serve as a neighborhood base at zero in a topological vector space:

Here are my questions:

Why "This shows that $\mathcal{N}$ is a neighborhood base at zero" and how this help to prove the uniqueness? I can understand the proof up to the the underscored sentence. But I don't see how to reach that point. (What is the construction of $A$ used for? How does the author show that $U\in\mathcal{N}$ is a neighborhood of zero?) 

The definition of a neighborhood base at some point in a topological space I know is as follows. 

 A: The definition of a neighbourhood base, which is used in the proof, seems to me to be different from the definition you posted. I think the proof uses the definition with the modification that we don't need $\mathcal N \subseteq \mathcal T$, but only that each set in $\mathcal N$ is a neighbourhood. This definition makes some sense, because we also distinguish neighbourhoods and open neighbourhoods (it's also the definition given by Wikipedia (https://en.wikipedia.org/wiki/Neighbourhood_system#Basis)).
In this case, indeed the passage shows that each $U \in \mathcal N$ is a neighbourhood of zero, and if $V$ is an arbitrary neighbourhood of zero, we choose an open set $O \subseteq V$, and then $0 + U \subseteq O$ for some $U \in \mathcal N$ by the definition of open sets.
Let now $\tau$ be a translation-invariant topology such that $\mathcal N$ is a neighbourhood base of zero.
Let $O \in \tau$. We choose some $x \in O$. Then $0 \in O - x \in \tau$ (translation invariance), and thus we find $V \in \mathcal N$ such that $V \subseteq O - x$, and then $V + x \subseteq O$. Hence, $O \in \mathcal T$, and since $O \in \tau$ was arbitrary, $\tau \subseteq \mathcal T$.
If $O \in \mathcal T$, then
$$
O = \bigcup_{x \in O} (x + U_x),
$$
where $U_x \subseteq V_x$ is open in $\tau$ and $V_x \in \mathcal N$ such that $x + V_x \subseteq O$. Hence, $O \in \tau$ as the union of open sets.
Thus, $\tau = \mathcal T$.
A: Prologue
Most common approaches to topology:
The Point-Set Opens, the Hausdorff Neighborhoods and the Kuratowsky Closure.
(It turns out that they are all equivalent approaches.)
They all come with their own advantages:
The Point-Set Opens approach is best when investigating purely topological properties like compactness or connectedness as it stays in the realm of subsets not diving into points.
The Hausdorff Neighborhoods approach fits best when relating distinct points topologically like in Hausdorff spaces or topological vector spaces as it defines neighborhoods w.r.t points.
The Kuratowski Closure approach untangles relations to hull operators of other categories like the linear span or the orthogonal complement as it is a selfmap on power sets.
Sometimes even a mixture is worthwhile like for locally compact spaces.
Definition
Given a plain space $\Omega$.
Consider $\mathcal{N}_x\subseteq\mathcal{P}(\Omega)$ for every $x\in\Omega$.
(We call $\mathcal{N}_x$ the neighborhoods of $x\in\Omega$.)
Every $\mathcal{N}_x$ should be a filter:
If $N$ and $N'$ are neighborhoods of $x\in\Omega$ then so is $N\cap N'$.
If $N$ is a neighborhood of $x\in\Omega$ then so is $A$ for every $N\subseteq A$.
Also $\mathcal{N}_x$ should be neighborhoods:
Itself $\Omega$ is a neighborhood for every $x\in\Omega$.
If $N$ is a neigborhood of $x\in\Omega$ then so for every $z\in N_0$.
(There $N_0$ is a small enough neighborhood of $x\in\Omega$.)
