Going from a fundamental system of neighborhoods to a topology and vice versa

Given for each $x \in X$ a nonempty set $U_x \subseteq \mathcal{P}(X)$ which satisfies
(1) $x \in \bigcap U_x$
(2) $V \in U_x, V \subseteq W \Longrightarrow W \in U_x$
(3) $V,W \in U_x \Longrightarrow V \cap W \in U_x$
(4) $\forall V \in U_x \,\exists W \in U_x: \forall y \in W: V \in U_y$
then there is exactly one topology on X s.t. for all $x \in X$: $U_x$ is the set of neighborhoods of $x$. A subset $O \subseteq X$ is open, iff it is a neighborhood of each of its elements - i.e. $\forall x \in O: O \in U_x$.

Both fundamental systems of neighborhoods, you mentioned above, are filter bases for the same $U_x$ which is obtained by adding all supersets in order to satisfy condition (2).

Your neighbourhood base $\mathcal V_x$ has to be non-empty and satisfy three properties:

(1) Each $V\in\mathcal V_x$ has to contain $x$.
(2) If $V_1,V_2\in\mathcal V_x$ then there is $V_3\in\mathcal V_x:V_3\subseteq V_1\cap V_2$.
(3) For each $V\in\mathcal V_x$ there is a $W\in\mathcal V_x$ such that for each $y\in W$ exists a $Y\in\mathcal V_y:Y\subseteq V$.

Then you can define $\mathcal U_x:=\{U\subseteq X\mid\exists V\in\mathcal V_x,V\subseteq U\}$ which satisfies (1) - (4) in Dune's answer. You can then proceed by defining a topology such that $\mathcal U_x$ is the neighborhood filter at $x$.

The $\left\{\left[x-\frac1n,\ x+\frac1n\right]\right\}_{n\in\mathbb{N}}$ neighborhood base would lead to the same neighborhood filter as $\left\{\left(x-\frac1n,\ x+\frac1n\right)\right\}_{n\in\mathbb{N}}$, since $\left[x-\frac1n,\ x+\frac1n\right]$ contains $\left(x-\frac1n,\ x+\frac1n\right)$, and $\left(x-\frac1n,\ x+\frac1n\right)$ contains $\left[x-\frac1{2n},\ x+\frac1{2n}\right]$.