The enlargement axiom does not trivialise the theory. It does imply that $X$ is a neighbourhood of every $x \in X$: every $x \in X$ has some $N \in N(x)$ (this is actually part of the axioms/conditions, often forgotten: every $N(x)$ is non-empty) and $N \subseteq X$ so $X \in N(x)$ by axiom 2. And everything inbetween $N$ and $X$ is a neighbourhood too. We cannot go smaller in general, so it's not true that everything becomes a neighbourhood of everything: $X$ is a neighbourhood of every point, but any fixed set $A$ is only a neighbourhood of $x$ if it happens to be the case that $A \in N(x)$, and otherwise it's not.
As to the necessity of the axiom: look at metric spaces, where $A$ is a neighbourhood of $x$ iff there is some $r > 0$ such that $B(x,r) \subseteq A$. The way this is defined already implies that larger sets than $A$ are also neighbourhoods of $x$; the same $r$ will work. So it's not a very surprising axiom. The same can be said for ordered spaces ($A$ is a neighbourhood of $x$iff either there exists $a \in X$ with $x < a$ and $\{z : z < a\} \subseteq A$, or there exists $b \in X$ with $b < x$ and $\{z : b < z \} \subseteq A$ or there exist $a,b \in X$ with $a < x, x < b$ and $\{z: a < z, z < b \} \subseteq A$), where also the enlargement is clearly true. We really are interested in the "small" neighbourhoods of $x$ (for continuity etc.) but in this axiomatisation we want to capture the notion of all possible neighbourhoods, even the not so interesting large ones. We could also axiomatise "basic" neighbourhoods instead, instead of all neighbourhoods, but this is not what we are doing here. In short, the relation of $A$ being a neighbourhood of $x$ is "local" and not affected by enlarging the set.
As to axiom 4, it is true that the "intended" result is that every neighbourhood of $x$ contains an open neighbourhood of $x$, but for that we first need to define what "open" means. We only have a set of sets $\{N(x): x \in X\}$ satisfying the axioms, no notion of openness. But if you already know the standard axioms for topology in terms of open sets, you want to go back and forth:
If you have a topology $\mathcal{T}$, define for each $x \in X$: $N(x) = \{A \subseteq X: \exists O \in \mathcal{T}: x \in O \subseteq A\}$ and check that this fulfills the axioms (again see that 2 is trivially fulfilled). Call this system $\mathcal{N}(\mathcal{T})$. On the other hand, if we have a system $\mathcal{N}= \{N(x): x \in X \}$ satisfying the axioms, define $\mathcal{T} = \{O \subseteq X: \forall x \in O: O \in N(x) \}$ and check that this defines a topology (we do not need axiom 4 for this, it turns out). Call this topology $\mathcal{T}(\mathcal{N})$. The axioms 4 is used to show that $\mathcal{N}(\mathcal{T}(\mathcal{N})) = \mathcal{N}$ again. So the neighbourhood system from the topology induced by the neighbourhood system is that same one we started with. It's a fun exercise to check this out in detail. It will enhance your understanding of these axioms. It's also true that $\mathcal{T}(\mathcal{N}(\mathcal{T})) = \mathcal{T}$, which is also fun to prove.
The usual way to define a topology $\mathcal{T}_d$ from a metric-like function $d: X \times X \to \Bbb R$ is to define $B_d(x,r)=\{y \in X: d(x,y) < r\}$ for $x \in X$ and $r>0$.
$O \subseteq X$ is then called open iff $$\forall x \in O: \exists r>0: B_d(x,r) \subseteq O\tag{1}$$
Checking the usual axioms for open sets:
$X$ is open is trivial, for any $x \in X$ we can take $r=1$ (or whatever) to fulfill $(1)$. $\emptyset$ is open because there are no $x$ in it to check $(1)$ on (void truth).
If $O_i, i \in I$ is a family of open sets, then $O=\bigcup_{i \in I}O_i$ is open: if $x \in O$, then for some $i_0 \in I$ we have $x \in O_{i_0}$. As that set is open by $(1)$ we have $r>0$ such that $B_d(x,r) \subseteq O_{i_0}$. Because $O_{i_0} \subseteq O$ (as always for unions) that same $r$ works to fulfil $(1)$ for $O$ and $x$. So $O$ is open.
If $O_1$ and $O_2$ is open, let $x \in O_1 \cap O_2$ be arbitrary. $x \in O_1$ gives us $r_1>0$ such that $B_d(x, r_1) \subseteq O_1$ and $x \in O_2$ gives us $r_2>0$ such that $B_d(x, r_2) \subseteq O_2$. Set $r=\min(r_1,r_2)$ and regardless of any axioms on $d$ we know that $$d(x,y) < r_1 \land d(x,y) < r_2 \iff d(x,y) < r$$ It follows that $$B_d(x,r) \subseteq O_1 \cap O_2$$ and $(1)$ is fulfilled for $x$ and $O_1 \cap O_2$. So $O_1 \cap O_2$ is open.
So this also defines a natural topology for a symmetric premetric $d$.
If $d$ is a full metric, we get the standard metric topology.
Best Answer
The empty set $\varnothing$ is not a neighborhood of any point $x\in X$, because as you correctly observed, there are no elements of $\varnothing$. However, the examples you are citing are not claiming that $\varnothing$ is a neighborhood, so there is no contradiction.
The examples are sets $X$ together with a topology $\tau$, so to say that a subset $U\subseteq X$ satisfies $U\in\tau$ means that $U$ is an open set, not (necessarily) a neighborhood of any particular point $x$.
From that Wikipedia page:
The subset $U=\varnothing$ vacuously satisfies this property, because there are no points in $\varnothing$.