Open Ball definition of Closure, Interior, and Boundary First the definitions in question. 
Closure: If $A$ is a subset of $\mathbb R ^n$, the closure of $A$ denoted $\bar{A}$, is the set of $x \in \mathbb R ^n$ such that for all $r > 0$ $$B_r (x) \cap A \ne ∅ $$
Interior: If $A$ is a subset of $\mathbb R ^n$, the interior of $A$ denoted $\text{int}(A) $ is the set of all $x \in \mathbb R ^n$ such  that there exists $r>0$ such that $$ B_r (x) \subset A $$
Boundary: The boundary of a subset $A \subset \mathbb R ^n$, denoted $\partial A$ consists of the points $x \in \mathbb R ^n$ such that every neighborhood intesects both $A$ and its compliment.
If we have "for all $r > 0 $", could we not pick an $x$ arbitrary outside of $A$ and make our $r$ large enough so that it still would still intersect with $A$? This same idea goes with the definition of boundary. We could pick a point far off into the compliment, and still make it intersect the set $A$ by choosing a large enough $r$. 
The equations relating the boundary, interior, and closure to each other make intuitive sense, but I can't see how it could make sense from these definitions. 
 A: To reiterate what Pedro had already said, sure you can pick $r$ large enough such that the $B_r(x) \cap A \neq \emptyset$, but this is not true for all $r$. In fact, you specified an $r$ -- a very large one.
Intuitively, you should think of the points in the closure $\overline{A}$ as points that have infinitely many points of $A$ around them so that no matter how small of a radius your ball is, you still have points from $A$.
A: i see you are having problems with the definition of closure:

Closure:
  given: $A \subseteq \mathbb{R^n}$ - the set closure is constructed as follows:
  $$
\bar{A}=\{x \in \mathbb{R^n}\mid \forall r \in \mathbb{R^+}:B_r(x)\ \cap A \neq \phi \}
$$

in the definition the inclusion must work for all $r$'s so it must work for very small one's, if you choose a point $\hat x$ "too far" from $A$ then there is a ball centered in $\hat x$ such that the intersection is empty! so the point is not in the closure! if you think about it only the points of $A$ and of the boundary of $A$ can satisfy that condition.  

more intuitively:
  a point $\bar x$ is in $\bar A$, iff $\bar x$ is surrounded by an infinite number of points of $A$ which are arbitrarily close to $\bar x$.

