My answer focuses on the AES matrix only. In general, an $A$ matrix with all submatrices having full rank generates an MDS code when concatenated by an identity matrix of the right size to form $[I|A]$, but that's really a coding theory matter.
As to the specific question, this is basic linear algebra but you need to work over the field that AES is defined over. Let the $k\times k$ matrix (like the AES mixing matrix) be
\vdots &&& \vdots \\
The $1\times1$ submatrices are all nonzero, they're just the entries of $A.$
The $2\times 2$ submatrices: choose any two rows or columns. you can also use the fact that the matrix is circulant to reduce the cases.
So the matrix
is the same matrix as
and thus has the same determinant, etc.
The $4\times$ submatrix is the whole matrix.
Row and column swaps only change the sign of the determinant and can be ignored.
Edit: You need to use the finite field with 256 elements as explained in the link below.
How to use the Extended Euclidean algorithm to invert a finite field element?
This is NOT mod 256 arithmetic.