Is the analogy of neighborhood as open ball applicable to arbitrary topological spaces? I've just started learning topology, and we frequently use examples of sets, open/closed subsets, etc. similar to this picture:

My question is that, in topological spaces where the open sets are not open balls, do neighborhoods $U_x$ of points x still have this property: 
A $\subset$ X, x $\in$ A
A is an open set <=> $\forall$ x $\in$ A, $\exists$ $U_x$ such that $U_x$ $\subset$ A
 A: Let $X$ be a topological space. We say that a set $N ⊆ X$ is a neighborhood of a point $x ∈ X$ if there is an open set $U ⊆ X$ such that $x ∈ U ⊆ N$.
It holds that a set $U ⊆ X$ is open if and only if it is a neighborhood of each of its points. This is also equivalent to the following condition: for every $x ∈ U$ there is a neighborhood $N_x$ of $x$ such that $N_x ⊆ U$.
Similarly, you can detect open set just by using basic sets. We say that a collection of open set $\mathcal{B}$ is an open base if every open set is a union of members of $\mathcal{B}$. It follows that a set $U$ is open if and only if for every $x ∈ X$ there is a basic open set $B_x ∈ \mathcal{B}$ such that $x ∈ B_x ⊆ U$. For a metric space, the collection of all open balls forms an open base of the topology – this is actually the definition of the topology induced by the metric.
Moreover, if you fix a point $x ∈ X$. An open base at $x$ is a collection $\mathcal{B}(x)$ of open neighborhoods of $x$ such that every neighborhood of $x$ contains a member of $\mathcal{B}(x)$, and usually it is also closed under finite intersections or at least directed. A local base is a collection $(\mathcal{B}(x): x ∈ X)$ of open bases at the respective points. Again, a set $U ⊆ X$ is open if and only if for every $x ∈ X$ there is a set $B_x ∈ \mathcal{B}(x)$ such that $B_x ⊆ U$. This structure is like an open base, but keeps the information which set comes from which point. In a metric space, we may consider $\mathcal{B}(x)$ to be the collection of all open balls centered at $x$.
