If $A$ and $B$ are sets of real numbers, then $(A \cup B)^{\circ} \supseteq A^ {\circ}\cup B^{\circ}$

Here is a slightly shorter proof:

We have $A^\circ \subset A \cup B$ and $B^\circ \subset A \cup B$, so we must have $A^\circ \cup B^\circ \subset A \cup B$.

Since $A^\circ \cup B^\circ$ is open and the interior is the largest open set contained in a set, we must have $A^\circ \cup B^\circ \subset (A \cup B)^\circ$.

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, $A^0\cup B^0 \subseteq (A\cup B)^0$.