Definition: For a subgroup $G$ of the group $PSL(2,\mathbb{C})$ acting on $\mathbb{P}^1$, its domain of discontinuity is the set of all points, $z$, with the following properties:
$1.$ The stabilizer $G_z$ of $z$ is finite.
$2.$ $\exists U$, a neighbourhood of $z$ such that,
$\space \space\space\space(a)\space $$\forall g\in G_z$, $gU=U$.
$\space \space\space\space(b)\space $$\forall g\in G - G_z$, $gU \cap U=\emptyset$.
Definition: A Kleinian group is defined to be a subgroup $G$ of $PSL(2,\mathbb{C})$ such that its domain of discontinuity is not empty.
Question: Consider the subgroup $\{M \in PSL(2,\mathbb{C}) \mid m_{ij}\in \mathbb{Z}[i], det {M}=1\}$. Why is this not Kleinian?
Is this clear from the above definition or would I have to prove some special property of Kleinian groups that this so called 'Picard Modular Group' does not satisfy?
The domain of discontinuity of this group is indeed empty. Hence by the definition of Kleinian group that you give this group is not Kleinian. That's an out-of-date definition, however; in most usages a Kleinian group is simply a discrete subgroup of $PSL_2(\mathbb{C})$, and then one uses the phrase "Type 2 Kleinian group" to refer to those which have nonempty domain of discontinuity.
Terminological evolutionary issues aside, for any discrete subgroup $G$ of $PSL(2,\mathbb{C})$, its action on $\mathbb{CP}^1$ subdivides that sphere into two sets: its domain of discontinuity which you have defined; and the complement of the domain of discontinuity called the limit set. One can characterize the limit set as the set of points of $\mathbb{CP}^1$ which are contained in the closure of the $G$-orbit of every point of $\mathbb{CP}^1$.
Your subgroup, also denoted $PSL_2(\mathbb{Z}[i])$, has limit set equal to the whole of $\mathbb{CP}^1$, and hence its domain of discontinuity is empty.
One way to prove this is consider the extended action to the upper half space $\{(x,y,z) \in \mathbb{R}^3 \mid z > 0\}$, thought of as hyperbolic 3-space $\mathbb{H}^3$, whose 2-sphere at infinity is identified with $\mathbb{CP}^1$. What one does is to construct a fundamental domain $D \subset \mathbb{H}^3$ for the action of $PSL_2(\mathbb{Z}[i])$ on $\mathbb{H}^3$. Then one computes the volume of $D$. It turns out that the volume of $D$ is finite. Then one invokes a theorem which says that if the volume of a fundamental domain of a discrete subgroup of $PSL(2,\mathbb{C})$ is finite then that subgroup has limit set equal to the whole of $\mathbb{CP}^1$.
I think there is most likely another way to prove this, using a little number theory. It turns out that the limit set of a discrete subgroup $G < PSL(2,\mathbb{C})$ is equal to the whole of $\mathbb{CP}^1$ if and only if the action of $G$ on $\mathbb{CP}^1$ has a dense orbit. I'm pretty sure that, when one identifies $\mathbb{CP}^1 = \mathbb{C} \cup \infty$, the orbit of $0=0+0i$ under the action of $PSL(2,\mathbb{Z}[i])$ is equal to $\mathbb{Q}[i] \cup \{\infty\}$, which is clearly dense in $\mathbb{C} \cup \infty$.