How is the name "open set" in topological space consistent with the intuitive definition of an open set? I can accept that "open set" is just a name used for the three characteristics of a topology in the definition.
Usually, when reading about open sets the definition goes like DEF := "an open set is a set that only includes its interior and not its boundary".
So, some questions:
a) Is this definition DEF just only true for metric spaces and does not have any correspondence to topological spaces?
b) If it does correspond to topological spaces, then how? For example, wikipedia has this image to describe what is a topology and what not: https://upload.wikimedia.org/wikipedia/commons/f/ff/Topological_space_examples.svg I do not even understand what the boundary or interior of 2 should include, what would that be?
c) If DEF does not hold for topological spaces, is "closed set" really just a name that has no further characteristics whatsoever?
 A: In general topological spaces, "interior" and "boundary" are defined such that the characterization you quote is still true -- but in that setting it doesn't work as a definition, because "interior" and "boundary" are now themselves defined in terms of open sets.
For metric spaces, a more direct definition would be

A subset $A$ of the metric space is called open if for every point $x\in A$, there is $\delta>0$ such that the ball $B_\delta(x)$ is wholly within $A$.

How does this relate to general topological spaces? What's going on is that whenever you have a metric space, then this definition is a standard way of defining a topology on the underlying set. In other words, metric spaces are the prototypical example of topological spaces, where "open" is defined using the metric.
But general topological spaces also allow cases where you have a class of subsets that are defined in a different way than by the above ball-based definition. If the family of sets you choose to call "open" satisfy the axioms for a topology, it will share enough properties with the open sets in a metric space that interesting parallels can be drawn.
A: For a topological space, your intuitive "definition" is true, since for any set $A$ we have $\partial A = \bar{A}-A^o$. If $A$ is open then $A = A^o$ and so $\partial A\cap A = \emptyset$. Conversely, if the last equality holds then $A^o = A-\partial A = A$.
Only problem is, the definition of the boundary already relies on the definition of an open set; so this definition for an open set is circular. In a metric space, a set is open if each of its elements have an $\epsilon$-ball around it contained in the set. Intuitively, these sets do not contain "boundary points", since any ball around a boundary point would contain points outside of the set. Then it turns out that these open  sets are closed under finite intersection and arbitrary union, and the whole space and $\emptyset$ are open. These last four requirements are then taken to be the only necessary properties of an open set in a topological space. 
A: An open set is an element of a collection s.t. $\mathcal C$ s.t. 
$$\bigcup_{O\in \mathcal A}O\in \mathcal C$$
for all $\mathcal A\subset \mathcal C$, and such that $$\bigcap_{i=1}^n O_i\in \mathcal C$$ for all $\{C_i\}_{i=1}^n\subset \mathcal C$.
Using this, you can then prove that in fact that $O$ is open iff $O=Interior(O)$. Notice that by definition, 
$$Interior(O)=\bigcup_{B\in \mathcal C:B\subset O}B.$$
