Set Theory - Elementary Proof I need someone to assess whether the proof I attempt to provide for the following statement is correct. 

Statement:
CORRECTED
$$ \left( A \cup B \right)\cap C \neq A \cup \left(B \cap C \right)$$

My approach: use indicator functions.

Two indicator functions $I_{S_1}$ are $I_{S_2}$ are identical if and
  only if the two sets, $S_1$ and $S_2$, are identical.

This means that $ \left( A \cup B \right)\cap C = A \cup \left(B \cap C \right)$ if and only if $I_{\left( A \cup B \right)\cap C} = I_{A \cup \left(B \cap C \right)}$.
$I_{\left( A \cup B \right)\cap C} = \left(I_A+I_B-I_A I_B\right)I_C = I_A I_C + I_B I_C - I_A I_B I_C$
$I_{A \cup \left(B \cap C \right)}= I_A + I_B I_C - I_A I_B I_C $
These 2 expressions are not always equal, therefore I conclude that these sets are not always identical.

I know, another approach I may try is to show that either $\left( A \cup B \right)\cap C \not\subset A \cup \left(B \cap C \right)$ or $\left( A \cup B \right)\cap C \not\supset A \cup \left(B \cap C \right)$. I am curious whether the first one above is rigorous enough?

I wrote wrongly the statement, which means some of the solutions provided below don't answer it precisely. Anyway I wasn't looking for a solution but rather for a proving method. Thanks anyone for suggestions!
 A: Just take $A = B = \{0\}$ and $C = \{1\}$. Then $A \cup B = A \cap B = \{0\}$ and
$$\left( A \cup B \right)\cap C = \{0\} \cap \{1\} = \{\} = \emptyset$$
while
$$\left( A \cap B \right)\cup C = \{0\} \cup \{1\} = \{ 0, 1 \}.$$
This simple examples shows that in general,
$$\left( A \cup B \right)\cap C \neq \left( A \cap B \right)\cup C.$$
There are cases where equality holds. Just take $A = B = C.$
A: The elements of $(A \cup B) \cap C$ must be in $C$ and the elements of $(A \cap B) \cup C$ might be in $C$.  Obviously these are different concepts and we can come up with a counter example but finding and element in the latter that is not in $C$.  
Say $1 \not \in C$ but $1 \in (A\cap B) \cup C$ so $1 \in (A\cap B)$.
So let $1 \in A; 1 \in B; 1\not \in C$.  Then $1 \not \in (A\cup B)\cap C \subset C$.  But $1 \in (A\cap B) \subset (A\cap B) \cup C$.  So $(A\cup B)\cap C\ne  C \subset (A\cap B) \cup C$.
If you want to be even more specific: Let $A = B = \{1\}$ and $C = \emptyset$.  Then $(A\cup B) \cap C = \emptyset$ while $(A \cap B) \cup C = \{1\}$.
A: You wrote down that
$$ \tag 1 I_A + I_B I_C - I_A I_B I_C \neq I_A I_C + I_B I_C - I_A I_B I_C  $$
but did not explain why the LHS is not always equal to the RHS. 
You need to finish off the argument with some simple observation. Well, if $C$ is the empty set, then $I_C = 0$ and (1) becomes 
$$ \tag 2 I_A  \ne 0$$
and I can think of some situations where (2) is true.
