Subset in Neighborhood Space is Neighborhood iff it contains Open Set

Theorem
Let $\left({S, \mathcal N}\right)$ be a neighborhood space.

Let $x \in S$ be a point of $S$.

Let $N \subseteq S$ be a subset of $S$.

Then $N$ is a neighborhood of $x$ there exists an open set $U$ of $\left({S, \mathcal N}\right)$ such that $x \in U \subseteq N$.

Necessary Condition
Let $N$ be a neighborhood of $x$.

Then by neighborhood space axiom $N 5$, $N$ contains a neighborhood $U$ of $x$ such that $U$ is a neighborhood of each of its points.

By neighborhood space axiom $N 2$, $x \in U$.

Sufficient Condition
Let $N$ be such that:
 * $\exists U \in \mathcal N: x \in U \subseteq N$

By definition of open set, $U$ is a neighborhood of $x$.

By neighborhood space axiom $N 3$, $N$ is then a neighborhood of $x$.