I'm not sure of what you want to know, but I'll try my best to explain: The Kronecker Delta can be introduced by the following relation: given the euclidean $n$-space with usual base $\left\{e_i\right\}$, then we have $\delta_{ij}= \left\langle e_i, e_j\right\rangle$, in other words, we define the Kronecker Delta as simply the inner product of the vectors from the base.
In truth, the inner product is bilinear, which means that if we have two vectors $v, w$, then we have their expansion in the usual base as being:
$$v=\sum_{i=1}^nv^ie_i$$
$$w=\sum_{j=1}^nw^je_j$$
Hence their inner product between those vectors is
$$\left\langle v,w\right\rangle = \left\langle \sum_{i=1}^nv^ie_i,\sum_{j=1}^nw^je_j\right\rangle$$
But since it's bilinear, we can take the sums out and the components of the vector, so that:
$$\left\langle v,w\right\rangle = \sum_{i=1}^n\sum_{j=1}^nv^iw^j\left\langle e_i,e_j\right\rangle$$
So, bilinearity implies that the inner product is completely specified if we define the inner product on the basis vectors, and so we simply set the Kronecker Delta as you've said:
$$\delta_{ij}=~\begin{cases}1 \hspace{3mm} \mbox{if} \hspace{3mm} i=j,\\
0 \hspace{3mm}\mbox{if} \hspace{3mm}i \neq j. \end{cases}$$
Because that's what we want that inner product to be: $e_i$ and $e_j$ should be orthogonal if $i\neq j$. Hence, introducing the Kronecker Delta is equivalent to introducing the usual inner product between two vectors. As I said, we build the Kronecker Delta to reflect the usual inner product completely. Also there's a relation with the Dirac Delta, but I think what you wanted to know is that. I hope it helps you.