Why is $\operatorname{Int}(A) \cup \operatorname{Int}(B) \neq \operatorname{Int}(A \cup B)$? I know that $\operatorname{Int}(A) \cup \operatorname{Int}(B) \subset \operatorname{Int}(A \cup B)$, but that the other direction does not hold, so can anybody please tell me whats wrong with the following proof? 
Suppose $x \in \operatorname{Int}(A \cup B) \Rightarrow \exists \epsilon > 0 : K(x, \epsilon) \subset A \cup B$. Assume without loss of generatlity that $K(x,\epsilon) \subset A-B \Rightarrow x \in \operatorname{Int}(A) \Rightarrow x \in \operatorname{Int}(A) \cup \operatorname{Int}(B)$.
 A: Because it's false, in general, that $\operatorname{Int}(A\cup B)\subseteq \operatorname{Int}(A)\cup \operatorname{Int}(B)$: take $A=(0,1]$ and $B=[1,2)$. Then $\operatorname{Int}(A)=(0,1)$, $\operatorname{Int}(B)=(1,2)$, but $\operatorname{Int}(A\cup B)=(0,2)$.
You can surely assume without loss of generality that $x\in A$, but not that the neighborhood $K(x,\epsilon)$ is contained is $A$. Try picturing what happens in the above example: $K(1,\epsilon)$ contains elements that are greater than $1$ (and belong to $B$) as well as elements that are less than $1$ (and belong to $A$). Since $A\cap B=\{1\}$, no neighborhood of $1$ can be contained either in $A$ or in $B$.

Consider a different case, but that should shed light over your difficulty: $X=\{0,1,2\}$, $Y=\{2,3,4\}$. Then $Z=\{1,3\}$ is a subset of $X\cup Y$, but $Z$ is not contained in any of $X\setminus Y$, $Y\setminus X$ or $X\cap Y$. It holds for elements that if $x\in X\cup Y$ than one of the following is true: $x\in X\setminus Y$, $x\in Y\setminus X$ or $x\in X\cap Y$.
A: If $X\subseteq A\cup B$, it need not be the case that $X\subseteq A$ or $X\subseteq B$.
As for a specific case, in the real numbers both the rationals and their complement have empty interior. What is the interior of the union? 
A: $K(x,\epsilon) \subset A \cup B$ does not implies $K(x,\epsilon) \subset A \Delta B$. It is not true when $K(x,\epsilon) \cap (A \cap B) \neq \phi$. For example, consider $A = (0,2)$ and $B = (1,3)$ and $x = 1.5$ and $\epsilon = 0.25$.
A: You assume without loss of generality that any point $x$ that lies in $A \cup B$ necessarily lies strictly in $A$ or strictly in $B$. If your point lies in the intersection $A \cap B$ your proof doesn't work. You have however proven the case where $A$ and $B$ are disjoint. 
An extension of your proof by saying that the ball $K(x, \varepsilon)$ lies entirely in the intersection will not work. Look for instance at the following example: Suppose we endow $\mathbb{R}$ with the Euclidean topology, then we can look at the intervals $[0,1]$ and $[1,2]$. it is clear that $$1 \in Int([0,2])$$ But we have $$ 1 \notin Int([0,1]) \qquad 1 \notin Int([1,2]) $$
Thus providing a counterexample.
A: $\operatorname{Int}([1,2]) \cup \operatorname{Int}([2,3]) =(1,2) \cup (2,3)$
$\operatorname{Int}([1,2] \cup [2,3]) = (1,3)$
$K(2, 1/10) \in \operatorname{Int}([1,2] \cup [2,3])$
There is no $\epsilon$ such that $K(2, \epsilon) \in [1,2]$ and $K(2, \epsilon) \in [2,3]$
