If $A$ and $B$ are sets of real numbers, then $(A \cup B)^{\circ} \supseteq A^ {\circ}\cup B^{\circ}$ I have a proof for this question, but I want to check if I'm right and if I'm wrong, what I am missing.
Definitions you need to know to answers this question: $\epsilon$-neighborhood, interior points and interiors.  Notation: $J_{\epsilon}(a)$ means a neighborhood formed around a (i.e. $(a-\epsilon, a+\epsilon)$.  An interior point in some set $A$ is a point where an $\epsilon$-neighborhood can be formed within the set.  The set of all interior points in $A$ is denoted as $A^0$ and is called the interior.
My proof for the question:  I'm proving based on most subset proofs where you prove that if an element is in one set, then it must be in the other.  Say $x \in A^0 \cup B^0$.  Then there is a $\epsilon > 0$, where $J_{\epsilon}(x) \subseteq A$ or $J_{\epsilon}(x) \subseteq B$.  This implies that $J_{\epsilon}(x) \subset A \cup B$ which then implies $x \in (A \cup B)^0$.  We can conclude from here that $(A \cup B)^0 \supseteq A^0 \cup B^0$.

Your proof is good- there is no problem with it. I have reworded a bit to make things a little more clear, specifically, where you say:

Then there is $\epsilon>0$, where $J_\epsilon (x)\subseteq A$ or $J_\epsilon (x)\subseteq B$.

Instead it would be more appropriately stated as: If $x\in A^0\cup B^0$ then $x\in A^0$ or $x\in B^0$. Then you can assume $x$ is in $A$, prove that $x\in (A\cup B)^0$, by symmetry the same argument will work if $x\in B^0$. 
I have included an edited proof- but as I originally stated your proof is good,
this just more directly reflects the definitions involved. 
Let $x\in A^0\cup B^0$. Then $x\in A^0$ or $x\in B^0$. Assume $x\in A^0$. Then there is, by definition, $\epsilon>0$ such that $J_{\epsilon}(x)\subseteq A$. Since $A\subseteq A\cup B$, we also have $J_{\epsilon}(x)\subseteq A\cup B$. Thus by definition we have $x\in (A\cup B)^0$. If instead $x\in B^0$, then by symmetry the same argument works. Thus, $A^0\cup B^0 \subseteq (A\cup B)^0$. 
