$ A = (A\cap E^c) \cup (A\cap E)$? $ A = (A\cap E^c) \cup (A\cap E)$ ?
is that correct for any sets A and E?
Here is my proof:
Let $x\in A$ since $A\subset A\cap E$ $\ x\in A\cap E \cup (A\cap E^c)$
now let $x\in (A\cap E^c) \cup (A\cap E)$ so either $x \in (A\cap E^c)$ or $x \in \cup (A\cap E)$ it follows that $ x \in A$ in both cases.
 A: That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.
$$
\begin{align}
A &=  A \cap U \hskip{1cm} \text{  (Identity law - $U$ is the universe)}\\
  &=   A \cap ( E \cup E^c) \hskip{1cm} \text{  (Complement law)}\\
  &=   (A \cap  E) \cup (A \cap E^c) \hskip{1cm} \text{  (Distributive property)}\\\\
\end{align}
$$
A: There is a flaw in your proof (if I understood it well). It is not true that $A \subset A \cap E$; for example, if $E = \{1,2\}$ and $A = \{0,1\}$, then $A \cap E = \{1\}$.

I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A \subset X$. 
Let $x \in A$. If $x \not\in E$, then $x \in E^c$, and hence $x \in A \cap E^c$.  Hence $A \subset (A \cap E)\cup(A \cap E^c)$.
To prove the other inclusion, note that both $(A \cap E)$ and $(A \cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.
If $A \not\subset X$ then you have $A \cap X = (A \cap E)\cup(A \cap E^c)$, since now $A \cap X \subset X$.
