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. 

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 $A$.

