Why are neighborhoods assumed to be open? I believe topology was defined in order to talk about limit points. In order to say things like 'point x is a limit point of set E' or, equivalently, 'x can be arbitrarily approximated by points in E'. The realization was that we don't actually need a metric. All we need is a notion of 'neighborhood', i.e. a little in every direction of a point. 
With the notion of neighborhood, we can now define limit point, interior, exterior, boundary, open set, closed set, etc. For example, a boundary point of a set E is a point x such that every neighborhood that contains x contains a point in E and a point not in E. An open set is a set those points are all interior points, and so on. 
All this seems intuitive. It makes sense that neighborhoods would be closed under arbitrary union and finite intersection, for example. What I do not understand is why the neighborhoods are assumed to be the open sets in a topology. Is there any reason behind this? 
 A: I think a neighbourhood space is what was originally used to formalise the idea of a neighbourhood. Here every point has a non-empty set of neighbourhoods $\mathcal{N}(x)$, such that


*

*Every $x$ is in every one of its neighbourhoods, or formally: $$\forall x : \forall N \in \mathcal{N}(x): x \in N$$

*Neighbourhoods are closed under intersections: $$\forall x \forall N_1, N_2 \in \mathcal{N}(x): N_1 \cap N_2 \in \mathcal{N}(x)$$

*Neighbourhoods are closed under enlargements: $$\forall x \forall N \in \mathcal{N}(x): \forall N' \subseteq X: (N \subseteq N') \to N' \in\mathcal{N}(x)$$ 
Note that 2 and 3 can be reformulated as saying $\mathcal{N}(x)$ is a filter on $X$ and 1 that is refined by the fixed filter on $x$. Also note that we only talk about "neighbourhood of $x$" not just "neighbourhood".
This set-up allows one to define interior point: $x$ is an interior point of $A$ iff $A \in \mathcal{N}(x)$ and from this the interior of a set $A$, the closure of $A$ etc., etc. A set is called open if it's its own interior, or $$\forall x \in O: O \in \mathcal{N}(x)$$ and this defines a topology on $X$. 
An open set is a neighbourhood of each of its points.
In this topology, being a neighbourhood of $x$ does not mean that that set is always open. E.g. $[-1,1]$ is a neighbourhood of $0$ in the usual topology but not open. If we take the set of all neighbourhoods of all points together (their union) then this is not closed under intersections: $[-1,1] \in \mathcal{N}(0)$, $[1,3] \in \mathcal{N}(2)$ but $\{1\} = [-1,1] \cap [1,3]$ is not a neighbourhood of any point, again in that topology. Open sets are special, because they are neighbourhoods of all its points.
Neighbourhoods in most texts are not assumed to be members of a topology, just that $N$ is a neighbourhood of $x$ if there is some open set $O$ just that $x \in O \subseteq N$. Open sets are neighbourhoods, but not reversely. I talked about why that can be convenient here. 
