Can one infer independence by simple reasoning/intuition? From my recent experience in probability, it feels as though independence is something we "discover" from the system via the equation: 
$$P(A)*P(B)=P(A\cap B)$$
Could one ever conclude independence from the "system" by intuition? Is it wise to conclude independence for events that are "seemingly" independent? What would be some interesting examples where this would fail. 
 A: The OP asks for an intuitive understanding of $\mathbb{P}(A)\mathbb{P}(B)=\mathbb{P}(A\cap B)$ for independent events $A$ and $B$. Here it is:
Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space. Assume $\mathbb{P}(B) > 0$. Since $\mathbb{P}(\Omega)=1$, we can write the above equation as
$\displaystyle\frac{\mathbb{P}(A)}{\mathbb{P}(\Omega)}=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}.$
The left hand side is the "proportion of $A$" in $\Omega$, while the right hand side is the "proportion of $A$" that is in $B$. From the equation above, we can see that the probability of the happening of $A$ remains unchanged inside $B$ (if you zoom in $B$ and treat it as a new probability space with $\mathbb{P}_B(B)=1$). Event $A$ will happen with a probability $\mathbb{P}(A)$ independent of whether your space is $\Omega$ or $B$. Thus is the independence of $A$ and $B$.
Imagine the $\Omega$ as a unit disk, and $A$ occupies the left semi-circle, so that $\mathbb{P}(A)=\frac{1}{2}$. Now let $B$ be a concentric circle inside $\Omega$. Then $A$ has exactly the same pattern in $B$ as in $\Omega$, and $\mathbb{P}_B(A)=\frac{1}{2}$ inside $B$. So in this respect $B$ is somewhat "irrelevant" for $A$: $B$ "looks like" $\Omega$, so its happening or not get unnoticed for $A$.

A: As @Masacroso has mentioned, independence in math is by definition the formula you've mentioned. If you're referring to real life systems, yes: Independence of events is a hypothesis that can be tested empirically - the relative frequency of a particular event may have been independent from that of another event. 
Is it wise to conclude independence empirically? Depends on the payoffs. If you're a decision-maker, you don't only care about being right or wrong; you care a lot too about the size of the payoff if you're right, and that of the penalty if you're wrong. It has been speculated, for instance, that a large contributor to the housing crisis was banks' assumption that mortgage failures of one household are independent of those of others. While this had been true in the past, it changed very quickly. The payoff from assuming independence was a relatively small gain from being able to make more loans; these gains and more were wiped out from the penalty for being wrong. (See Nassim Taleb's books for related discussion.)
A: An interesting example is as follows
Suppose there are two fair six faced dice, say I and II. Both dice are rolled together, let $X$ denotes the face shown by die I and $Y$ denotes the face shown by die II. 
Define $Z:= X+Y$ and $U := $ remainder when $Z$ is divided by $6 $.
Surprisingly in the above construction random variables $U$ & $X$ are independent. 
For example, let $A= \{X= 1\}  $  & $B=\{ U= 1\} = 
\{ (X,Y)= (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\} $
Then $\mathbb{P}(A)= \dfrac{1}{6}$,  $\mathbb{P}(B)= \dfrac{1}{6}$ and $\mathbb{P}(A\cap B)= \mathbb{P}(\{ (X,Y)= (1,6)\})=  \dfrac{1}{36}= \mathbb{P}(A) * \mathbb{P}(B)$.
