The "standard" definition of closure that I usually encounter is the dual definition of interior. One defines interior as the largest open set that is contained in $A$; this is achieved with the definition you give (the union of all open sets contained in $A$) since the union of open sets is open, and the union of subsets of $A$ is a subset of $A$.
To achieve the dual definition, we define the closure of $A$ to be the smallest closed set that contains $A$, which is then easily seen to be equivalent to: $\mathrm{cl}(A) = \bigcap\{ C\mid A\subseteq C\quad\text{and}\quad C\text{ closed}\}.$ (Or, one can define the closure of $A$ to be the complement of the interior of the complement of $A$, $\mathrm{cl}(A) = \left(\mathrm{int}(A^c)\right)^c,$ since the smallest closed set that contains $A$ is the complement of the largest open set that is contained in $A^c$).
Now, you want to prove that \mathrm{int}(A)\cup \partial A = A\cup A'.
To prove \mathrm{int}(A)\cup\partial A \subseteq A\cup A', note that \mathrm{int}(A)\subseteq A \subseteq A\cup A' by definition, so it suffices to show that \partial A \subseteq A\cup A'. Take $x\in\partial A$. If $x\in A$, there is nothing to do. Now assume $x\notin A$; use the definition of $\partial A$ to show that you will necessarily have x\in A'.
The converse inclusion is perhaps a bit more delicate, since you don't have any obvious inclusions. I would proceed as follow: first show that $A\subseteq \mathrm{int}(A)\cup \partial A$, by showing that if $a\in A-\mathrm{int}(A)$, then $A\in\partial A$; to do this, show that any open set containing $a$ cannot be completely contained in $A$. Then we need to show that A'\subseteq \mathrm{int}(A)\cup\partial A. Show that if x\in A'-A, then $x\in \partial A$. Then simply note that if x\in A'\cap A, then you're done already, since you already know that $\mathrm{int}(A)\cup \partial A$ contains every point of $A$.
Finally, we probably want to check that the "standard" definition I gave agrees with the ones you were given.
If $x\notin \mathrm{int}(A)\cup\partial A$, then $x\notin\partial A$, so there is an open set that contains $x$ and is either completely contained in $A$, or completely contained in $A^c$; but if it were completely contained in $A$, then we would have $x\in\mathrm{int}(A)$, a contradiction to the choice of $x$; thus, there is an open set $U$ with $x\in U$ and $U\cap A=\emptyset$. That means that $C=U^c$ is closed and contains $A$, so $\mathrm{cl}(A)\subseteq C$. Since $c\notin C$, then $x\notin \mathrm{cl}(A)$. We have shown that $\mathrm{cl}(A)\subseteq \mathrm{int}(A)\cup\partial A$.
Now let $x\notin \mathrm{cl}(A)$. Then there exists a closed set $C$ such that $A\subseteq C$ and $x\notin C$. Let $U=C^c$; then $U$ is open, $U\cap A=\emptyset$, and $x\in U$. In particular, $x\notin A$, and x\notin A', so x\notin A\cup A'. Thus, we have shown that A\cup A'\subseteq \mathrm{cl}(A).
Hence: A\cup A' \subseteq \mathrm{cl}(A) \subseteq \mathrm{int}(A)\cup\partial A = A\cup A', proving the desired equality.